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259 Geometry Writings by
Christopher J.Bradley full pdf articles, separate pages with titles and summaries
09 -02-2017
collected by Chronopoulos Takis (parmenides51) Obituary *
CHRISTOPHER J. BRADLEY (1938–2013) Christopher Bradley was a popular and effective Fellow and Tutor in Mathematics at Jesus College from 1964 to 1977, after which he entered schoolteaching, notably at Clifton College, near Bristol. As Deputy Leader of the British Mathematical Olympiad team and for many years Secretary of the British Mathematical Olympiad Committee, he was known for his elegant Olympiad problems. He was also involved with the UK Maths Trust and the Mathematical Association. While in Oxford he co-authored a major study of the mathematical theory of symmetry in solids, and later wrote introductory texts for the UK Maths Trust and a book on Challenges in geometry: for mathematical Olympians past and present. The following problems are taken from one of Christopher Bradley’s books: Prove that one member of a Pythagorean triple is always divisible by 5, and that the area of any right-angled triangle with integer sides is divisible by 6. * taken from: https://www.maths.ox.ac.uk/system/files/legacy/3277/newsletter_0414_0.pdf Contains separate pages with all titles and all summaries. For your own convenience use the bookmarks on the left of this pdf. Articles’ Source: http://people.bath.ac.uk/masgcs/bradley.html parmenides51 facebook
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Christopher Bradley's Geometry Writings
Titles 1 The Story of Hagge and Speckman 2 The four Hagge Circles 3 Generalizations of Hagge's Theorem 4 Studies in Similarity 5 Omega circles 6 On the Ten point Wood-Desargues' Configuration 7 Generalization of the Wallace-Simson line. 8 A singular Miquel configuration and the Miquel direct similarity 9 Miquel circles and Cevian lines 10 A new construction to find any circle through a given point. 11 Eight Circles through the Orthocentre. 12 Circles concentric with the Circumcircle 13 Ex-points and Eight Intersecting Circles 14 Intersecting Circles having Chords the sides of a Cyclic Quadrilateral 15 Four concurrent Euler Lines 16 The Direct Similarity of the Miquel Point Configuration 17 Harmonic Ranges in a Coaxal system of Circles 18 Some Special Circles in a Triangle 19 Circular Perspective 20 On the Nine Intersections of two Cevian Triangles 21 Significant Points on Circles Centre the Circumcentre 22 Six Points on a Circle 23 The Symmedian point and the Polar Line 24 The Thirteen Point Circle 25 When Quadrangles are completely in Perspective 26 A Circle concentric with the Incircle 27 Porisms with a circular circumconic 28 Some circles in a Cyclic Quadrilateral 29 The Miquel Circles for a Quadrilateral 30 More on Circular Perspective 31 More cases of Circular Perspective 32 The GH Disc and anogther case of Triple Circular Perspective 33 29 Circles, 34 Some circles in the Cyclic Quadrilateral configuration 35 Where 7 circles meet Part 1 36 Where 7 circles meet Part 2
37 More Circles in the Cyclic Quadrilateral Configuration 38 Eight Points on a line and Seven Circles through a Point 39 The 60°, 75°, 45° triangle and its Euler line 40 The 120°, 45°, 15° triangle and its Euler line 41 The 105°, 60°, 15° Triangle and its Euler Line and Circles 42 A Cyclic Quadrilateral, 29 Points and 33 Lines 43 The Altitudes and Radii of a triangle and its circumcircle 44 A natural Enlargement resulting in a Collineation 45 On the Circles defining the Brocard points 46 A Mean of two Cevian points and the Construction of Triangle centres 47 A Cevian point and its six harmonics 48 The Symmedian point and its Harmonics 49 Perpendicular Bisectors and Angle Bisectors 50 Three triangles in mutual triple perspective 51 When one Conic produces two more 52 Three triangles in mutual Triple Reverse Perspective 53 Hagge Circles touching at H 54 Four Triangle Conics 55 Perpendiculars to a Triangle's sides through its Vertices 56 The Cevian point Conic 57 Perpendiculars to the Cevians at the Cevian Point 58 Additional results for the Miquel Configuration 59 7 Points on any Circle not through a Vertex 60 Centroid-centred similar ellipses 61 Concurrent lines in a triangle with a Circle cutting the Sides 62 A Triangle with an arbitrary Conic cutting its Sides 63 A Cascade of Conics 64 Porism constructed by the Circumcircle and Triangles in Perspective 65 When I replaces K and Ge replaces H and Mi replaces O 66 On the mean of two Cevian points 67 A Special Tucker Circle 68 Six Collinear Points in a Special Cyclic Quadrilateral 69 Two Cyclic Quadrilaterals and Two Coaxal Systems
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Christopher Bradley's Geometry Writings 70 What happens when a Triangle is Rotated about its Orthocentre 71 More Special Cyclic Quadrilaterals 72 What happens when you reflect a Triangle in any Line 73 Constructing Two Coaxal systems by Incidence and Reflection 74 The Simson Line Porism 75 The Double Simson Line Circle 76 The Double Simson Line Conic 77 The Circle centre O and radius OH 78 More on the Seven-Point CircleMore on the Seven-Point Circle 79 Conics in the Ex-circle Configuration 80 Incircle and Excircle Conics 81 A Radical Centre that lies on OI 82 Cevian derived Conics 83 Conics generated by Points on a Curve of degree Five 84 The G Circles and the Conic they determine 85 A Converse of the Pascal Line Property 86 The Perpendiculars to Three Segments at a Point determine a Conic 87 Two connected Rectangular Hyperbolae 88 A Singular Cyclic Quadrilateral 89 Perpendiculars in a Cyclic Quadrilateral 90 Perpendiculars from the vertices of a Cyclic Quadrilateral 91 6 Conics 92 Three special Cyclic Quadrilaterals 93 Circles formed by an Isosceles Trapezium 94 Generalization of the Steiner Point 95 A Nine Point Rectangular Hyperbola 96 More on the Nine Point Rectangular Hyperbola 97 How the Excentres create Points on the Circumcircle 98 Ex-points and their Sets of Intersecting Circles 99 A Triangle and its Image under a Half Turn 100 Circles though a point in an Equilateral Triangle 101 Intersections at the vertices of the Second Brocard Triangle 102 Two Triangles, their Ex-Symmedians and four Conics 103 The Altitudes Create Four Circles, Four Conics and a Polar line
104 How any Six Points on a Circle create Two Conics 105 External Squares on the Sides of a Triangle 106 On Perpendiculars at the corners of a Cyclic Quadrilateral 107 Conics through three points with Centres on a Fixed Line 108 A Property of 3 Circles passing through a fixed Point 109 Four Perspectives given a Triangle and its Circumcircle 110 Three Concurrent Pascal lines and an Auxiliary Conic . 111 Perpendicular bisectors of Three Radii of a Circle . 112 Parallels from the feet of Cevians create a Conic . 113 Circles passing through the Miquel Point of the feet of Cevians 114 Properties of a Particular Tucker Circle 115 Six Circles and their Centres . 116 The Geometry of the Brocard Axis and Associated Conics 117 When the Cevians of triangle ABC meet the Circumcircle at D, E, F 118 Two Triangles in Perspective and inscribed in a Conic 119 The Nine-point Conic and a Pair of Parallel Lines 120 An Important Line through the Centre of a Cevian Inellipse 121 Some Lines through the Centroid G 122 Some lines through the Incentre I 123 Properties of the Intangential Triangle 124 Two Cevian Points Collinear with a Vertex and Thirteen Conics 125 Properties of the Extangential Triangle 126 Some Properties of the Ex-Circle configuration 127 When a Circle passes through a Vertex and cuts the Opposite Side 128 The Symmetry of a Scalene Triangle 129 The Neuberg Cubic 130 An Extension to the Theory of Hagge Circles 131 Construction of Circles always having centre the Nine-Point Centre 132 The Cevian Conic
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Christopher Bradley's Geometry Writings 133 Conics in a semi-regular hexagon 134 When the Cevian Conic is a Circle 135 Constructing Triangles with Coincident Centroids 136 Properties of the symmedian point in a cyclic quadrilateral 137 Affine tranform of the properties of the Eulerian triangle 138 Explaining some collinearities among triangle centres 139 The harmonic cevian conic 140 The centroid of centroids 141 Three centroids created by a cyclic quadrilateral 142 Incircle conjugation 143 The nine-point circle of the diagonal point triangle 144 Four nine-point circles 145 Concyclic circumcentres in the Steiner configuration 146 More conics in a semi-regular hexagon 147 The midpoint rectangular hyperbola again 148 A circle through two vertces, three circumcentres and a Miquel point 149 Circles with a common point in a cyclic quadrilateral 150 When a point is a circumcentre, an incentre and an orthocentre 151 The Orthocentre Conics 152 Generating Circles from the Symmedian point 153 More properties of the Incentre 154 On the Incircle and Excircles of a Cevian Triangle 155 The Circumconic of a pair of Cevian Triangles 156 Circumcentre Conics 157 The General Inellipse 158 Six Point Circles and their Associated Conics 159 A Twelve Point Configuration and Carnot’s Theorem 160 Tangents to a Conic from the vertices of a Triangle 161 When 12 points display eight 6-point Conics and 6 Concurrent Lines 162 When 24 Points form three 8-point Conics and 12 Concurrent Lines 163 A Theorem on the Complete Quadrilateral 164 The Transversal of a Quadrilateral
165 A description of properties of Pascal's hexagon 166 Analytic treatment of a Romanian problem 167 Basic Properties of a Quadrangle possessing an Incircle 168 Properties of a pair of Diametrically Opposite Triangles 169 Two In-Perspective Triangles inscribed in a Conic 170 The Brocard Conics 171 A Conic through the feet of two Cevians may lead to a second Conic 172 The Inner Circle of a Cyclic Quadrilateral 173 The Auxiliary Circles of a Cyclic Quadrilateral 174 A Cevian Circle leads to Perspective Triangles 175 When two Triangles intersect in Conics and in Circles 176 Orthologic Triangles 177 Special Pascal Lines 178 More Circles centred on the Brocard Axis 179 The Second Brocard Triangle 180 Circles through the Brocard points and the Circumcentre 181 Circles through the Brocard points and the Symmedian Point 182 Problems requiring Proof 183 Conics and Triangles in Perspective 184 Cevian Perspectivity 185 The Four Conic Theorem 186 6 Points and 4 perspectives 187 A Perspective in a pair of Cyclic Quadrilaterals 188 Are three Triangles ever in Mutual Perspective? 189 The Midpoint Conic 190 The Brocard Circles' Twin Conic 191 Three Circles and their Centres 192 Generalization of the Triplicate Ratio Circle 193 A Two Circle Problem Involving the GK axis 194 The KG Nine-Point Conic 195 The Orthocentre Circle of a Cyclic Quadrilateral 196 The Two Central Lines in a Cyclic Quadrilateral 197 When Two Pairs of Diagonals are Concurrent 198 The Two Cevians' Perspective 199 Three Circles with collinear Centres 200 Problem on Quadrilateral with an Incircle 201 All about a Cyclic Quadrilateral and its cousin 202 The Midpoint Theorem 203 Another Circle with Centre on the Brocard axis
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Christopher Bradley's Geometry Writings 204 Between the Triplicate Ratio Circle and the Circumcircle 205 The Brocard Lines KΩ and KΩ' 206 Dividing OH into Five or Seven Equal Parts 207 The Median Conic 208 A Pair of Isogonal Conjugates produce a Conic through 6 Points 209 Three Concentric Circles 210 12 Points, 8 six-point Conics, 4 Conics through each Point 211 A cyclic quadrilateral and its midpoint circles 212 Quarter size Circles at Triangle Vertices 213 When an Incircle produces a Circumcircle 214 The Super-Cevian Triangles, their Conic and Two Perspectives 215 Multiplication of Points using Barycentric Coordinates 216 A Point, two Triangles and Two Conics 217 An interesting Perspective in the Anticomplementary Triangle 218 Properties of the Triangle of Excentres 219 The Twelve point Circle 220 A Conical Hexagon with Main Diagonals Concurrent 221 Four Special Conical Hexagons all with same Polar 222 The Missing Point on the Euler Line 223 When a non-regular Cyclic Pentagon leads to Another 224 The Sixty Pascal Poles 225 A Typical Pascal line Drawing 226 Special Conical Hexagons 227 The Remarkable Eight-Point Conic 228 How 2 Conics through 4 Points generate 6 more such Conics 229 When a Conical Quadrilateral produces two 6point Conics 230 A Quadrilateral and the Conics and Points arising 231 A Quadrilateral and resulting Conics Part 2 A Special Case showing the role of the New Points X and Y 232 The Coconic Hexagon with Main Diagonals Concurrent 233 The Conic and the Lines that are Created (Part 1) 234 A Conic generates five 8-point Conics (Part 2)
235 The Truth and the whole Truth about a Quadrilateral 236 A Nice Conic and a Nasty Circle 237 How to Construct the Ex-points of a given Point 238 When Two Quadrilaterals are in Complete Perspective 239 More on Complete perspective 240 Pentagons in Complete Perspective 241 Quadrilaterals in Perspective inscribed in a Conic 242 Perspective Quadrilaterals 243 Some properties of a Special Hexagon 244 When two Triangles in Perspective create Conics 245 Some Simple Algebraic Results 246 Pascal points 247 The G Conic 248 The H Conic 249 Twin Conics 250 Cevian Circles 251 The Four Brocard Triangles 252 Odds and Ends 253 Triangles in Perspective – the Whole Truth 254 Circles through the midpoints of sides of a Cyclic Quadrilateral 255 Five Conics and a Polar line 256 Two Quadrangles in Perspective in a Conic 257 Quadrangles in Perspective Part II The three lines 258 The Miquel Cyclic Quadrilateral 259 Two Triangles and an Eight Point Conic
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Christopher Bradley's Geometry Writings Titles and summaries Article 1 is The Story of Hagge and Speckman . This concerns theory first developed by Hagge and Speckman in the Edwardian era. Speckman investigated triangles which were simultaneously in perspective, and indirectly similar. On the other hand Hagge studies circles which pass through the orthocentre of a given triangle. Superficially these subjects look unrelated, but this is not the case. Article 2 is The four Hagge Circles . This describes properties of the four Hagge circles of triangles BCD, ACD, ABD, ABC when ABCD is a cyclic quadrilateral. Article 3 is Generalizations of Hagge's Theorem . Two possible generalizations of Hagge circles are considered. The first describes those properties preserved when the circle passes through a point other than the orthocentre. The second when a pair of orthologic triangles are involved. Article 4 is Studies in Similarity . The six points: the orthocentre, the Brocard points and the points where the medians intersect the orthocentroidal circle involve nine circles through the vertices of ABC and either the orthocentre or one of the Brocard points. The triangles formed by the centres of these circles exhibit many similarities. Article 5 is Omega circles . It is shown how circles through either of the Brocard points have properties similar to Hagge circles. Article 6 is On the Ten point Wood-Desargues' Configuration . The ten pairs of directly similar triangles in perspective in the Wood-Desargues' configuration have ten common Hagge circle centres. It is shown in this paper that these centres lie at the vertices of five cyclic quadrangles, which are similar to the five cyclic quadrangles of the initial Wood-Desargues' configuration. Article 7 is Generalization of the Wallace-Simson line. In this article it is shown how a direct similarity transformation of triangle ABC into another triangle enables one to identify the Double Wallace-Simson line of the new triangle as a generalized Wallace-Simson line of ABC. Article 8 is A singular Miquel configuration and the Miquel direct similarity. In the Miquel configuration for a triangle the centres of the three concurrent circles form a triangle similar to ABC. It is shown how the direct similarity between the two triangles involved produces further results of significance. Article 9 is Miquel circles and Cevian lines. The Miquel configuration has interesting additional properties when the points on the sides of ABC are the feet of Cevian lines. Article 10 is A new construction to find any circle through a given point. This is another generalization of the Hagge construction when the point involved is not the orthocentre. Article 11 is Eight Circles through the Orthocentre. Article 12 is Circles concentric with the Circumcircle. It is shown that such circles contain seven points with special properties. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings Article 13 is Ex-points and Eight Intersecting Circles. Points internal to a triangle have three points external to the triangle associated with them. Properties of these Ex-points are investigated. Article 14 is Intersecting Circles having Chords the sides of a Cyclic Quadrilateral. The circles pass through the centre of the cyclic quadrilateral or the intersection of its diagonals. Article 15 is Four concurrent Euler Lines. In this paper we consider a cyclic quadrilateral PQRS in which the diagonals PR and QS meet at a point E and we prove a number of results about the triangles PQE, QRE, RSE, SPE. Article 16 is The Direct Similarity of the Miquel Point Configuration . This continues work begun in Article 8 and is best read in conjunction with it. Article 17 is Harmonic Ranges in a Coaxal system of Circles. A circle cuts a triangle in six points and the condition that the harmonic conjugates of those points also lie on a circle is investigated. The coaxal system involving the polar circle, the circumcircle and the orthocentroidal circle feature. Article 18 is Some Special Circles in a Triangle. There is one special circle through each point in the plane of ABC, not on the sides or the circumcircle. Eight points on each circle have similar properties to those in Hagge circles. Article 19 is Circular Perspective. Two triangles ABC and UVW are in circular perspective when circles AVW, BWU, CUV all pass through the same point. It is shown that this concept is symmetric and two triangles in double circular perspective are automatically in triple circular perspective. Article 20 is On the Nine Intersections of two Cevian Triangles. A Cevian triangle is one whose sides join the three feet of a Cevian point. Two Cevian triangles intersect in nine points which exhibit a vast number of properties. Article 21 is Significant Points on Circles Centre the Circumcentre. Given a triangle ABC with circumcentre O and a point P not on its sides or their extensions and not on the circumcircle, it is shown that one may construct on the circle centre O and radius OP six significant points. Article 22 is Six Points on a Circle. A construction with a triangle ABC and the point X76 is described, yielding a circle bearing a distinct similarity to the Triplicate Ratio Circle. Article 23 is The Symmedian point and the Polar Line. A construction based on the Symmedian point is described yielding six lines through each of three points on the polar line. Two related porisms are constructed. Article 24 is The Thirteen Point Circle. In this article we give an account of the properties of the coaxal system of circles passing through the two Brocard points and having the Brocard axis as line of centres.
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Christopher Bradley's Geometry Writings Article 25 is When Quadrangles are completely in Perspective . In this article we establish a sufficient condition for when a pair of quadrangles have what may be appropriately called a Desargues' axis of perspective. Article 26 is A Circle concentric with the Incircle. A construction is given in a triangle using Gergonne's point yielding a circle concentric with the incircle. Article 27 is Porisms with a circular circumconic. The result we prove in this article is if we take the orthocentre of all the triangles in the porism in a case when the circumconic is the circumcircle, then the path traced out by the orthocentre is circular (or linear). Article 28 is Some circles in a Cyclic Quadrilateral. The properties of some circles in a cyclic quadrilateral passing through its vertices, midpoints of sides and centre are investigated. Article 29 is The Miquel Circles for a Quadrilateral. It is shown how these may be constructed starting from the Miquel points of triangles involving the diagonal points. Article 30 is More on Circular Perspective. It is shown how a triangle may be in circular perspective with three points on a line (rather than a second triangle). An example involving the intersections of the tangents at the vertices of ABC with the opposite sides is given. Article 31 is More cases of Circular Perspective . This article involves triangles ABC and PQR, where P, Q, R are the intersections of the medians with the orthocentroidal circle. Article 32 is The GH Disc and anogther case of Triple Circular Perspective, and continues the work on triple circular perspective, this time in connection with the orthocentroidal circle. Article 33 is 29 Circles, and continues the work on exploring triple circular perspective. Article 34 is Some circles in the Cyclic Quadrilateral configuration. When ABCD is a cyclic quadrilateral, centre O, and BA^CD = E, BC^AD = F and AC^BD = G, then circles ABF, CDF, BCE, ADE meet at a point T on EF. The perpendicular from T to EF passes through G and O. The centres of the four circles, O and T lie on a circle. Article 35 is Where 7 circles meet Part 1. The orthocentroidal circle S of triangle ABC on GH as diameter possesses an unusual property. If you draw circles BHC, CHA, AHB to meet S again at points X, Y, Z and circles BGC, CGA, AGB to meet S again at U, V, W then the following property holds: circles AYX, BZX, CXY, AVW, BWU, CUV all pass through a point Q on the circumcircle of ABC. Article 36 is Where 7 circles meet Part 1. The result of Article 35 is generalized to involve an arbitrary circle. Article 37 is More Circles in the Cyclic Quadrilateral Configuration. Quadrilateral ABCD, circle centre O has external diagonal points E and F. Circles on AO, CO as diameters meet at Y and circles on BO, DO as diameters meet at Z. It is shown that T, Y, Z are collinear, where T is the midpoint of EF. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings Article 38 is Eight Points on a line and Seven Circles through a Point. Points D, X, Y, Z and E, U, V, W lie on a transversal of triangle ABC. It is shown what the condition is connecting D and E for circles ABC, AYZ, BZX, CXY, AVW, BWU, CUV to meet at a point. Article 39 is The 60°, 75°, 45° triangle and its Euler line. Key results concerning the intersection of the Euler line and the sides of the triangle are established. Article 40 is The 120°, 45°, 15° triangle and its Euler line. Many results involving the triangle, its Euler line and various circles are established. Article 41 is The 105°, 60°, 15° Triangle and its Euler Line and Circles. Some remarkable properties of the 105°, 60°, 15° triangle and its intersections with its Euler line are investigated. Article 42 is A Cyclic Quadrilateral, 29 Points and 33 Lines. It is shown that the quadrilateral formed by the symmedian points of triangle BDC, ACD, ABD, ABC of a cyclic quadrilateral ABCD has the same diagonal point triangle as the cyclic quadrilateral ABCD. Article 43 is The Altitudes and Radii of a triangle and its circumcircle. In triangle ABC, circumcentre O and orthocentre H if ab is the intersection of AH and OB, then the midpoints of pairs ab, ac and ba, bc and ca, cb lie on the circle on OH as diameter and circles ab bc ca and ba ac cb both pass through H. Article 44 is A natural Enlargement resulting in a Collineation . This is a short survey of the properties of the Exsimilicentre X56 and shows that it is on the line joining Feuerbach point and the Orthocentre. This is a known collineation. Article 45 is On the Circles defining the Brocard points . A construction is given in which points X, Y, Z lie on a circle containing the first Brocard point and points X', Y', Z' lie on a circle containing the second Brocard point. The six circles such as BXC are the Brocard circles and the six circles such as AYZ always pass through Tarry's oint. Article 46 is A Mean of two Cevian points and the Construction of Triangle centres . A construction is given which gives rise to about 9 million more triangle centres. It is part serious and part jocular, but its aim is to question the activity of hunting triangle centres. Article 47 is A Cevian point and its six harmonics If L is the foot of a Cevian point on BC then B1 is the harmonic conjugate of L and C and C1 is the harmonic conjugate of L and B. When similar points on CA and AB are defined a configuration is produced with many interesting features. Article 48 is The Symmedian point and its Harmonics Results of Article 47 are applied when the Cevian point is the Symmedian point. Article 49 is Perpendicular Bisectors and Angle Bisectors When these are drawn for a given triangle the six non-trivial points of intersection create a figure with many significant results
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Christopher Bradley's Geometry Writings Article 50 is Three triangles in mutual triple perspective . A configuration with pairs of points that are isotomic conjugates produces a figure with three triangles in mutual triple perspective. Six of the perspectrices pass through a given point. Article 51 is When one Conic produces two more . ABC is a triangle, circumcentre O. The given conic is ABCLMN, where L, M, N lie on AO, BO, CO. Twelve points are then formed from these points which are shown to lie 6 by 6 on two conics. Article 52 is Three triangles in mutual Triple Reverse Perspective . A construction and supporting analysis is given that ensures the perspectives. When two of the triangles have vertices on the circumcircle the Brocard porism is established and the third triangle degenerates and becomes the polar of the Symmedian point with respect to the circumcircle. Article 53 is Hagge Circles touching at H . A construction of two triangles is described whose circumcircles turn out to be a pair of touching Hagge circles. One of the triangles has a vertex at H and an interesting connection with the nine-point centre is established. Article 54 is Four Triangle Conics . The construction of the Triplicate Ratio Circle and the 7-Point Circle when the generating point is not the Symmedian point is described. Analysis shows that four (and not just two) Conics emerge. Those that might be termed the Triplicate Ratio Conic and the 7-Point Conic are similar and have the same centre. Article 55 is Perpendiculars to a Triangle's sides through its Vertices Six points on the sides are generated (other than the feet of the altitudes). If one constructs the three circles through B, C and C, A and A, B that cut the circumcircle orthogonally, the six points are the intersections of these circles with the sides of ABC. Article 56 is The Cevian point Conic In triangle ABC let P be a Cevian point with D, E, F the feet of the Cevians on BC, CA, AB respectively. Points L, M, N lie on AD, BE, CF respectively and are such that P is the midpoint of AL, BM, CN. It is found that the points A, B, C, L, M, N lie on a conic which we call the Cevian point conic of P. Several other properties are established. Article 57 is Perpendiculars to the Cevians at the Cevian Point The points where these perpendiculars meet corresponding sides are shown to be collinear. Article 58 is Additional results for the Miquel Configuration The Miquel point P is the common point of circles AMN, BNL, CLM when L, M, N lie on BC, CA, AB respectively. If perpendiculars to BC at L, CA at M, AB at N are drawn, producing six points on the other sides of ABC, then a variety of results hold. In particular three new circles may be drawn, pairs of which each have a common point with one of the Miquel circles. Article 59 is 7 Points on any Circle not through a Vertex Three Miquel circles AMN, BNL, CLM with L, M, N on the sides meet at a point P. S is any other circle through P (not through a vertex). AMN meets S at G and X, BNL meets S at G and Y, CLM meets S at G and Z. It is proved that AX, BY, CZ concur at a point lying on S. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings Article 60 is Centroid-centred similar ellipses The three ellipses are the outer and inner Steiner ellipses and an intermediate on passing through points a quarter and three quarter away along the sides. Certain sets of collinear points are identified. Article 61 is Concurrent lines in a triangle with a Circle cutting the Sides A circle cuts the sides of a triangle at the feet of two Cevian points. When this construction is carried out a number of other sets of three lines are concurrent and sets of three points are collinear. Article 62 is A Triangle with an arbitrary Conic cutting its Sides A conic is chosen that cuts the sides BC, CA, AB of a triangle ABC in points L, U; M, V; N, W (in that order anticlockwise) respectively. The chords LW, MU, NV form a triangle DEF. A large number of concurrencies and collinearities are identified. Article 63 is A Cascade of Conics A pair of in-perspective triangles are inscribed in a conic. Their noncorresponding sides meet at six points, which turn out to lie on a second conic and form a pair of inperspective triangles. The process may therefore be repeated indefinitely forming the cascade of conics in the title. Article 64 is Porism constructed by the Circumcircle and Triangles in Perspective A hexagon is formed by the tangents to the circumcircle at the vertices of the two triangles and a conic passes through the vertices of the hexagon. Article 65 is When I replaces K and Ge replaces H and Mi replaces O An analogue of the triplicate ratio circle and the 7-point circle is created under these circumstances (Mi is the Mittelpunkt and Ge is Gergonne's point). Article 66 is On the mean of two Cevian points The feet of their Cevians are used in a construction of a mean that has a geometrical significance. Article 67 is A Special Tucker Circle The Tucker circle passes through the feet of the perpendiculars to other two sides from the foot of the altitude through their common vertex. The properties of the resulting hexagon are reviewed. Article 68 is Six Collinear Points in a Special Cyclic Quadrilateral The cyclic quadrilateral has its diagonals at right angles and a review is given of the main properties of such a quadrilateral. Article 69 is Two Cyclic Quadrilaterals and Two Coaxal Systems It is shown how two cyclic quadrilaterals inscribed in the same circle centre O, each with their diagonals at right angles, generate four circles whose centres lie two by two on lines through O. In each case a coaxal system is generated. Article 70 is What happens when a Triangle is Rotated about its Orthocentre This is concerned with the properties of two triangles related by a rotation.
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Christopher Bradley's Geometry Writings Article 71 is More Special Cyclic Quadrilaterals This article is concerned with a cyclic quadrilateral ABCD in which AC is perpendicular to BD and its image A'B'C'D' after a rotation of 90o about E = AC^BD. Article 72 is What happens when you reflect a Triangle in any Line A triangle ABC and its circumcircle are mapped by a reflection line L with image triangle A'B'C' and its circumcircle. What happens is that the line through A' parallel to BC, the line through B' parallel to CA and the line through C' parallel to AB are concurrent at a point P that always lies on circle A'B'C'. Article 73 is Constructing Two Coaxal systems by Incidence and Reflection The cyclic quadrilateral ABCD, centre O, with diagonals AC and BD at right angles at E is reflected in the line AC to form the cyclic quadrilateral A' B'C'D', centre O'. From this configuration two systems of coaxal circles are obtained. Article 74 is The Simson Line Porism Let ABC be a triangle and let A'B'C' be its image under a 180 degree rotation about the circumcentre O. Now let D be any point on the circumcircle S. The locus of the point of intersection of the Wallace-Simson lines of D with respect to the two triangles is an inconic of both triangles, creating a porism. Article 75 is The Double Simson Line Circle ABC is a triangle and A'B'C' is its image under a rotation by 180 degrees about the circumcentre O. With P a point on the circumcircle the Double Simson lines of P with respect to the triangles are drawn, meeting at a point Q. The locus of Q as P moves round the circumcircle turns out to be the circle centre O. Article 76 is The Double Simson Line Conic A transversal LMN of a triangle is drawn through the orthocentre, which is the Double Simson line of a point P on the circumcircle. The corresponding Cevian point of the harmonic conjugates of L, M, N is a point Q, whose locus as P moves is a circumconic of ABC. Article 77 is The Circle centre O and radius OH A number of related results are established about the circle centre the circumcentre O of a triangle ABC and radius OH, where H is the orthocentre of triangle ABC. Article 78 is More on the Seven-Point CircleMore on the Seven-Point Circle Various properties are established about the Brocard circle and the circles BKC. CKA, AKB through the symmedian point K of a triangle ABC. Article 79 is Conics in the Ex-circle Configuration From the points of contact of the three ex-circles of a triangle ABC four conics are drawn and their properties are investigated. Article 80 is Incircle and Excircle Conics A configuration of four conics passing through the points of contact of the incircle and excircles of a triangle ABC is investigated. The intersections and centres of these conics define sets of points whose joins provide sets of concurrent lines.
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Christopher Bradley's Geometry Writings Article 81 is A Radical Centre that lies on OI The configuration of a triangle ABC with incentre I and the three circles BCI, CAI, ABI is investigated and three new circles are defined whose radical centre is a point J on OI where O is the circumcentre. J is shown to be the isogonal conjugate of Gergonne’s point. Article 82 is Cevian derived Conics Lines through the feet of Cevians of a point are drawn parallel to the sides of a triangle and meet the sides in six points lying on a conic. Properties of this conic are investigated. Article 83 is Conics generated by Points on a Curve of degree Five From the feet of Cevians of a point P lines are drawn perpendicular to the sides meeting them in six point. It turns out that these six points lie on a conic if, and only if, P lies on a curve of degree five whose equation is determined. Well known points lying on this curve are O, H, I, the circumcentre, orthocentre and incentre of ABC, and also the excentres. Article 84 is The G Circles and the Conic they determine From the feet of Cevians of a point P lines are drawn perpendicular to the sides meeting them in six point. It turns out that these six points lie on a conic if, and only if, P lies on a curve of degree five whose equation is determined. Well known points lying on this curve are O, H, I, the circumcentre, orthocentre and incentre of ABC, and also the excentres. Article 85 is A Converse of the Pascal Line Property It is shown how any transversal of a triangle and any three additional points lying one on each side of the triangle (but not at the vertices) may be used to find six points on the sides of the triangle through which a conic always lies. The proof involves a particular case of the converse of the Pascal line theorem. Article 86 is The Perpendiculars to Three Segments at a Point determine a Conic It is shown how any transversal of a triangle and any three additional points lying one on each side of the triangle (but not at the vertices) may be used to find six points on the sides of the triangle through which a conic always lies. The proof involves a particular case of the converse of the Pascal line theorem. Article 87 is Two connected Rectangular Hyperbolae Given a Cyclic Quadrilateral and the Rectangular Hyperbola Σ through the midpoints P, Q, R, S of its sides and its centre O, it is shown that this conic Σ automatically passes through the diagonal points. A slight extension of this famous result is documented. Article 88 is A Singular Cyclic Quadrilateral When a Cyclic Quadrilateral ABCD is such that the tangents at A and C and the line BD are concurrent, it follows that the tangents at B and D and the line AC are also concurrent. Additional properties of this configuration are obtained. Article 89 is Perpendiculars in a Cyclic Quadrilateral. Perpendiculars from A and C on to opposite sides of a cyclic quadrilateral ABCD produce four more cyclic quadrilaterals and two sets of parallel lines one set containing five lines and the other set containing six lines. Article 90 is Perpendiculars from the vertices of a Cyclic Quadrilateral . The perpendiculars from A to AB and DA and the six similar perpendiculars from B, C, D to the sides of a cyclic quadrilateral create two more cyclic quadrilaterals that are coaxal with ABCD. Their sides and their diagonal point lines exhibit some surprising properties. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings Article 91 is 6 Conics . When a circumconic is drawn to triangle ABC the tangents at A, B, C form a triangle DEF which is in perspective with ABC. If the lines AD, BE, CF meet at U and meet the circumconic at R, S, T then the tangents at R, S, T form a triangle LMN, It is shown that D, E, F, L, M, N lie on a conic, Then, by drawing tangents and chords four other conics may be constructed with centres all lying on a line through U. Article 92 is Three special Cyclic Quadrilaterals . In any cyclic quadrilateral ABCD if AB and CD meet at F and AD and BC meet at G then FG is one side of the diagonal point triangle. It is then always the case that the tangents at B and D and the tangents at A and C meet at points U and V respectively lying on FG. The first special cyclic quadrilateral is when AC passes through U. If the tangents at A and B meet at P and Q, R, S are similarly defined then the second special cyclic quadrilateral is when AC is perpendicular to BD and then P, Q, R, S are concyclic. The third special cyclic quadrilateral is when AC is parallel to BD and then again P, Q, R, S are concyclic. Article 93 is Circles formed by an Isosceles Trapezium . Given an isosceles trapezium ABCD with AD parallel to BC, if tangents to the cyclic quadrilateral ABCD are drawn to produce six points of intersection, then six more circles may be drawn, all passing through the centre O of ABCD. Five of these circles are obvious but the fact that the sixth circle passes through O is an interesting result and in this paper a proof is given using Cartesian co-ordinates. Article 94 is Generalization of the Steiner Point . The outer Steiner ellipse passes through A, B, C and the images L, M, N of those vertices in a rotation of 180 degrees about the centroid G. The Steiner point is the fourth point of intersection of the outer Steiner ellipse and the circumcircle of ABC. A generalization is obtained by replacing G by another point P internal to the triangle ABC. But more occurs as circles AMN, BMN, CNL intersect at a point U that lies on both the circumcircle and the ellipse ABCLMN and U is the generalized Steiner point. Article 95 is A Nine Point Rectangular Hyperbola . The configuration consisting of the rectangular hyperbola passing through the incentre, the excentres, the centroid and deLongchamps point in a triangle and three other significant points exhibits some interesting properties which are investigated. Article 96 is More on the Nine Point Rectangular Hyperbola . The circumcircle of a triangle ABC is the Nine-Point circle of the triangle IJK of its excentres. The line JK passes through A and the midpoint D of JK, which lies therefore on the circumcircle. With E, F similarly defined there are thus two triangles ABC and DEF and it is shown that their incentres and excentres lie on the nine point rectangular hyperbola discussed in CJB/2010/95. Other interesting properties emerge such as the deLongchamps point of DEF is the same point as the incentre of ABC. Article 97 is How the Excentres create Points on the Circumcircle . The circles through pairs of vertices of a triangle and the excentres opposite the third vertex have centres lying on the circumcircle and pass through the incentre of the triangle. The triangles with these centres as vertices exhibit properties that are described. Article 98 is Ex-points and their Sets of Intersecting Circles . The ex-symmedian points and the ex-points of an arbitrary point internal to the triangle are shown to lie on a hyperbola. The configuration involving Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings ex-points of a given point and vertices of a given triangle is shown to produce a set of three circles having a point in common (similar to the Fermat point). The relationship between these points is obtained. Article 99 is A Triangle and its Image under a Half Turn . A triangle ABC and its image DEF under a half turn results in circles BCD, CAE, ABF intersecting at a point P on circle DEF. A point Q on circle ABC is similarly defined. Also it is established that a conic passes through A, B, C, D, E, F, P, Q. Article 100 is Circles though a point in an Equilateral Triangle . Given three points D, E, F lying on the medians of an equilateral triangle ABC the conditions are determined for the circles BCD, CAE, ABF to have a common point. Article 101 is Intersections at the vertices of the Second Brocard Triangle . If D, E, F are the exsymmedian points it is proved that circles BCD, CAE, ABF intersect the Brocard circle at the circumcentre of ABC and at the vertices of the second Brocard triangle. A case of circular perspective is thereby determined. Article 102 is Two Triangles, their Ex-Symmedians and four Conics . The two triangles are in perspective, vertex the centroid of one of them. The six ex-symmedian points lie on a conic, the tangents at the vertex circumscribe the circumcircle and also the intersections of various lines produce six points lying three by three on two parallel lines. Two other conics of interest are also created. Article 103 is The Altitudes Create Four Circles, Four Conics and a Polar line . If the altitudes of a triangle ABC meet the circumcircle at D, E, F, then the six interior intersections of the two triangles and the tangents at the above six points produce a configuration in which four circles and four conics feature. An analysis is given in which the polar of the orthocentre plays a major role. Article 104 is How any Six Points on a Circle create Two Conics . Given three chords of a circle, perpendiculars to those chords from their end points create three pairs of parallel lines. Of their fifteen points of intersection three lie on the line at infinity and the other twelve lie six by six on two conics. Note that the six points must lie on a circle and not a general conic. Article 105 is External Squares on the Sides of a Triangle . If squares are drawn externally on the sides of a triangle the configuration exhibits two points similar to the Fermat points. Also the six intersections of various lines define a circle. Article 106 is On Perpendiculars at the corners of a Cyclic Quadrilateral . If perpendiculars to the sides are drawn at the vertices of a cyclic quadrilateral then a configuration is created that consists of two circles coaxal with the first circle, each containing quadrilaterals similar to each other. Their external diagonal points also lie three by three on two collinear lines. Two other circles also emerge. Article 107 is Conics through three points with Centres on a Fixed Line . Conics through three noncollinear points A, B, C having their centres on a fixed line have three radical axes, BC, CA, AB. If one takes a fixed point P on one of these axes, say BC, then the polars of P with respect to all the conics pass through another fixed point P', which as one would expect, is the harmonic conjugate of P with respect to B and C. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings Article 108 is A Property of 3 Circles passing through a fixed Point . Let ABC be a triangle and P a fixed point not on the sides. Circle BPC meets AB at W and AC at M. Circle CPA meet BC at U and BA at N. Circle APB meets CA at V and CB at L. The perpendicular bisectors of LU, MV, NW meet at a point Q. Article 109 is Four Perspectives given a Triangle and its Circumcircle . Given a triangle, its circumcircle and a point not on the circumcircle, it is shown how by varying the position of the point along a line four perspectives may be created. Article 110 is Three Concurrent Pascal lines and an Auxiliary Conic . Article 111 is Perpendicular bisectors of Three Radii of a Circle . Article 112 is Parallels from the feet of Cevians create a Conic . From the feet of Cevians through a point parallels to the other two sides are drawn. The six points created on the sides of the triangle lie on a conic. Article 113 is Circles passing through the Miquel Point of the feet of Cevians . If LMN are the feet of a set of Cevians, then circles AMN, BNL, CLM meet at a Miquel point Q. If now circles BQC, CQA, AQB are drawn and circles AMN and BQC meet at R with S, T similarly defined, then circle RST passes through the Miquel point Q. The centres of the seven circles also exhibit some remarkable properties showing Q to be the Miquel point of a second triangle, a cascade process that can be carried on indefinitely. Article 114 is Properties of a Particular Tucker Circle . Three circles determine a particular Tucker circle and their centres exhibit some interesting properties. Article 115 is Six Circles and their Centres . In triangle ABC centroid G, circles BGC, CGA, AGB are drawn meeting the sides in six points that lie on a conic. Three more circles are drawn and the six circle centres exhibit some interesting properties. Article 116 is The Geometry of the Brocard Axis and Associated Conics . From a triangle and its symmedian point K and its circumcentre O, various triangles, the Brocard inellipse and two further conics are constructed having their centres on the Brocard axis KO. Two porisms are defined and a special line is identified. Most, if not all, of the content of this article is known and indeed should be known by all geometers. Article 117 is When the Cevians of triangle ABC meet the Circumcircle at D, E, F . In a triangle ABC when the Cevians through the centroid G meet the circumcircle Σ at points D, E, F and the tangents are drawn at the six points, two triangles are created. The intersections of the sides of the four triangles in the configuration have some interesting properties. Article 118 is Two Triangles in Perspective and inscribed in a Conic . Given two triangles ABC, DEF in perspective at P and inscribed in a conic Σ, a configuration is produced consisting of the 6 Pascal lines and 2 Steiner points arising. One of the Pascal lines has further properties, as it is not only a Pascal line, but also a Desargues’ axis of perspective and also the polar of P with respect to Σ. Further properties involving the nine lines AD, AE, AF, BD, BE, BF, CD, CE, CF and the tangents at the six vertices are
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Christopher Bradley's Geometry Writings deduced leading to a conic, further properties of the key Pascal line and numerous sets of collinear points and concurrent lines. Article 119 is The Nine-point Conic and a Pair of Parallel Lines . An affine transformation of the ninepoint circle leads to the nine-point conic. The orthocentre transforms into a general point P which is collinear with the centre of the nine-point conic and the centroid. We prove that the polar of P with respect to the nine-point conic is parallel to the Desargues’ axis of perspective of the triangle and the image of the orthic triangle. Article 120 is An Important Line through the Centre of a Cevian Inellipse . A Cevian inellipse is one that touches the sides of a triangle internally at points that are the feet of Cevian lines. It is shown in this short article that the line through the Cevian point and the centre of a Cevian inellipse contains other significant points. Article 121 is Some Lines through the Centroid G . Using areal co-ordinates we catalogue some lines through the centroid G of a triangle ABC. Co-ordinates of points and equations of lines are given. Article 122 is Some lines through the Incentre I . Using areal co-ordinates we catalogue some lines through the incentre I of a triangle ABC. Co-ordinates of points and equations of lines are given. Article 123 is Properties of the Intangential Triangle . Triangle ABC and its incircle and ex-circles are given. Common tangents to the incircle and an ex-circle are four in number; three being the sides of ABC and the fourth is called an intangent. There are three intangents, one for each ex-circle and these form a triangle A'B'C' called the intangential triangle. Properties of the intangential triangle are studied using areal co-ordinates with ABC as triangle of reference. By construction its incircle is the same as that of triangle ABC. The triangle of its excentres has sides parallel to those of ABC. Article 124 is Two Cevian Points Collinear with a Vertex and Thirteen Conics . Given triangle ABC, a point P on BC and lines LMP and UVP with L, U on AB and M, V on CA, then two Cevian points may be constructed collinear with vertex A. Two further Cevian points occur and the resulting configuration of points and lines results in the appearance of thirteen conics and two harmonic ranges. Article 125 is Properties of the Extangential Triangle . The external common tangents to the three excircles form the extangential triangle A'B'C'. Triangles ABC and A'B'C' are in perspective with perspector the orthocentre of the intouch triangle. The axis of perspective plays a prominent part in the configuration. Triangle A'B'C' and the orthic triangle are homothetic through the Clawson point. Article 126 is Some Properties of the Ex-Circle configuration . The ex-circle configuration is studied and it is shown in particular how the midpoints of the sides of the ex-central triangle lead to several circles and perspectives, as well as a new (to me) triangle with vertices on the sides of the original triangle and two new (to me) points on the circumcircle. Article 127 is When a Circle passes through a Vertex and cuts the Opposite Side . A surprising configuration emerges when a circle passes through a vertex and cuts the opposite side. Two pairs of touching circles and two circles each passing through seven key points emerge and additional points on Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings the original circle are also created. It seems unlikely that a synthetic solution will exist, as some of the coordinate working is technically very difficult. Article 128 is The Symmetry of a Scalene Triangle . Points dividing each side of a scalene triangle in a fixed ratio leads to a configuration with two sets of three equal circles and a conic. Many results are selfevident but the equations of the circles and conic are recorded. Article 129 is The Neuberg Cubic . Analysis is made of the Neuberg triangle cubic. This is the locus of points P with the property that the reflection of P in the sides of a triangle ABC leads to a triangle which is in perspective with ABC. Article 130 is An Extension to the Theory of Hagge Circles . An extension of the theory of Hagge circles is presented. Article 131 is Construction of Circles always having centre the Nine-Point Centre . From any general point P reflections in BC. CA, AB give D, E, F. The midpoints of AD, BE, CF are U, V, W respectively. U, V, W are shown to lie on a circle with centre the Nine-Point centre. Article 132 is The Cevian Conic . If P is a Cevian point in a triangle ABC and D, E, F are the feet of the Cevians, then three circles PEF, PFD, PDE meet the sides again in six other points. It is shown that these six points lie on a conic, which we call the Cevian Conic. Article 133 is Conics in a semi-regular hexagon. Hexagon AFBDCE with ABC an equilateral triangle and AD, BE, CF concurrent at the centroid P of ABC is inscribed in a conic, and as such is defined as a semiregular hexagon. It is proved that the six circumcentres of triangles AFP, FBP, BDP, DCP, CPE, EPA are co-conic as are their six orthocentres. It is also proved that the intersections of contiguous Euler lines are co-conic. Cabri indicates that the six in-centres and six nine-point centres are also co-conic. Article 134 is When the Cevian Conic is a Circle . When the Cevian point is Gergonne's point, the Cevian Conic is a Circle with centre at the incentre. Article 135 is Constructing Triangles with Coincident Centroids . Given two triangles with coincident centroids it is shown how to construct an unlimited number of triangles with the same centroid. Article 136 is Properties of the symmedian point in a cyclic quadrilateral. If ABCD is a cyclic quadrilateral then the symmedian points of triangles ABC and DBC are shown to have some remarkable properties. Many pairs of lines may be constructed whose intersections lie on the sides of the diagonal point triangle. Article 137 is Affine tranform of the properties of the Eulerian triangle. Given a triangle ABC and a point K a configuration is constructed containing a triplicate ratio conic, a circumconic, a nine-point conic and a 7 point conic. When K is actually the symmedian point K(a2 b2, c2), then these conics are the familiar circles associated with a triangle in the Euclidean plane. In this article we take K, a point not on the sides of ABC, to be an arbitrary point K(f, g, h) and what emerges is an affine map of the Euclidean plane.
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Christopher Bradley's Geometry Writings Article 138 is Explaining some collinearities among triangle centres. Collinearities are preserved by an Affine Transformation, so that for every collinearity of triangle centres, there is another when an affine transformation has taken place and vice-versa. We illustrate this by considering what happens to three collinearities in the Euclidean plane when an affine transformation takes the symmedian point into the incentre. Article 139 is The harmonic cevian conic. If D, E, F are the feet of Cevians with D on BC etc. and if L is the harmonic conjugate of C with respect to B and D and U is the harmonic conjugate of B with respect to C and D and if M, N, V, W are similarly defined then a conic passes through the six points L, U, M, V, N, W. There is one internal Cevian point for which the conic is a circle. Article 140 is The centroid of centroids. Let P be a point not on the sides of a triangle and suppose the centroids of triangle PBC, PCA, PAB are L, M, N respectively. Now let J be the centroid of triangle LMN. Then J lies on line PG, where G is the centroid of triangle ABC. The point of concurrence Q of AL, BM, CN also lies on PG. Article 141 is Three centroids created by a cyclic quadrilateral . The centroid of the quadrilateral considered to be an area of constant density is G, the centroid of the quadrilateral considered as having unit masses at its vertices is N, the centroid of the quadrilateral considered as having unit masses at its vertices and mass of two units at E (the intersection of its diagonals) is F. It is shown that E, F, N, G are collinear. Article 142 is Incircle conjugation . A construction is presented in which points (f, g, h) in barycentric coordinates are transformed into points (a/f, b/g, c/h). This is clearly a conjugation and we call it incircle conjugation. Article 143 is The nine-point circle of the diagonal point triangle . An analysis is conducted of the ten point rectangular hyperbola through the mid-points of a cyclic quadrilateral and of the nine-point circle of the diagonal point triangle. Article 144 is Four nine-point circles . In any quadrilateral ABCD the four nine-point circles of triangles BCD, ACD, ABD, ABC have a point in common that lies on the hyperbola through the mid-points and on the circumcircle of the diagonal point triangle. Article 145 is Concyclic circumcentres in the Steiner configuration . In triangle ABC with centroid G, the points D, E, F lie on AG, BG, CG respectively and are such that AG = GD, BG = GE, CG = GF. It is proved that the circumcentres of triangles AFG, FBG, BGD, DGC, CGE, EGA are concyclic. Article 146 is More conics in a semi-regular hexagon . ABC is an equilateral triangle and AFBDCE is a hexagon inscribed in a conic. Triangles ABC and DEF are in perspective with vertex P, the centroid of ABC. G1-G6 are the centroids of triangles EAF, FBD, DCE, BDC, CEA, AFB. It is shown that these points lie on a conic. Also triangles G1G2G3 and G4G5G6 are congruent and in perspective. The six interior points of intersection of triangles ABC and DEF also lie on a conic.
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Christopher Bradley's Geometry Writings Article 147 is The midpoint rectangular hyperbola again . If A, B, C, D are four points on a circle and E is any other point, then the locus of the centre of conic ABCDE as E varies is the rectangular hyperbola through the midpoints of the quadrangle ABCD. Article 148 is A circle through two vertces, three circumcentres and a Miquel point . Starting with a triangle ABC and a point D on BC it is shown how the circumcentres of triangles ABC, ADB, ADC lie on a circle through A. If this circle intersects circle ABC at A and P, then it is shown that triangle PBC exhibits a Miquel point Q with points D, M, N on its sides. Article 149 is Circles with a common point in a cyclic quadrilateral If ABCD is a cyclic quadrilateral it is possible to find points P, Q, R, S, T, U on sides AB, BC, CD, DA, AC, BD respectively so that circles BPQU, APST, CQRT, DRSU have a common point M. These circles have centres that are concyclic, as are points M, U, I, T where I = AC^BD. Article 150 is When a point is a circumcentre, an incentre and an orthocentre . Given a triangle ABC, circumcentre O, and the centres D, E, F of circles BOC, COA, AOB respectively, then O is the incentre of triangle DEF, as well as being the orthocentre of the triangle of excentres. Article 151 is The Orthocentre Conics . When circles BHC, CHA, AHB in a triangle ABC with orthocentre H are drawn their centres D, E, F have interesting properties. Article 152 is Generating Circles from the Symmedian point . Given a triangle ABC and its symmedian point K circles BKC, CKA, AKB are drawn. Their other points of intersection with the sides of ABC are shown to lie on a Tucker circle. These points also lie in pairs on the sides of another triangle and the other intersections of these sides with the circles BKC, CKA, AKB are also concyclic. Centres of these circles are shown to lie on the Brocard axis. Article 153 is More properties of the Incentre . In a triangle ABC, with incentre I, the midpoints of AI, BI, CI are located. Perpendicular bisectors of AD, BE, CF are drawn. These meet the interiors of the sides of ABC at six points that are shown to lie on an ellipse. As the figure develops to include the excentres several other circles and conics are located. Article 154 is On the Incircle and Excircles of a Cevian Triangle . Given triangle ABC, if D, E, F are the feet of the Cevians AD, BE, CF and L, M, N are the points of contact of the incircle of triangle DEF with the sides of triangle DEF then AL, BM, CN are con ncurrent, as are AL', BM', CN', where L', M', N' are the points of contact of the ex-circles with EF, FD, DE respectively. Article 155 is The Circumconic of a pair of Cevian Triangles . In triangle ABC two Cevian triangles are drawn and their points of intersection and the vertices of ABC lead to six sets of collinear points. Several conics are identified. Article 156 is Circumcentre Conics . Given triangle ABC, let triangle UVW be a reduction of ABC by an enlargement factor t (0 < t < 1) about the circumcentre O. The sides of triangle UVW cut the sides of ABC in six points that lie on a conic. It is shown that the centres of all such conics lie on the Brocard axis.
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Christopher Bradley's Geometry Writings Article 157 is The General Inellipse . The general inellipse in a triangle touches its sides at the feet of Cevians. An analysis is given that relates the Cevian point to the centre of the inellipse and to two related circumconics. Article 158 is Six Point Circles and their Associated Conics . If P is a point internal to a triangle ABC and AP, BP, CP meet the circumcircle again at A', B', C' respectively and U, V, W are the midpoints of AP, BP, CP and D, E, F are the midpoints of PA', PB', PC' respectively then U, V, W, D, E, F are obviously concyclic. But the configuration has other properties which are pointed out. Article 159 is A Twelve Point Configuration and Carnot’s Theorem . A twelve point configuration occurs when six of the points lie on a conic and the remaining six points are the vertices of two triangles in perspective. In this article we consider the configuration in which a conic cuts the three sides of a triangle in real points. A number of theorems create a variety of conditions on the points of intersection equivalent to Carnot’s theorem. Article 160 is Tangents to a Conic from the vertices of a Triangle . Given a conic which cuts the sides of a triangle in six real points, tangents A1, A2, B3, B4, C5, C6 are drawn to the conic. The tangents meet each other at 12 points (15 if you count A, B, C). The se form a 12 point configuration (see Article 159) so that six of the twelve points lie on a conic. Article 161 is When 12 points display eight 6-point Conics and 6 Concurrent Lines . Six lines being tangents to a circle at vertices of triangles in perspective create twelve points of intersection. Six conics with six points each can be drawn through the twelve points, each point lying on three of the conics. Also six lines can be drawn through the twelve points, lines which are concurrent at the vertex of perspective. Article 162 is When 24 Points form three 8-point Conics and 12 Concurrent Lines . A construction is described involving two chords of a conic that results in 24 points that form three 8-point Conics and 12 concurrent Lines. Article 163 is A Theorem on the Complete Quadrilateral . Given a complete quadrilateral ABCDEF and two points P and Q then it is well-known that conics ABCPQ, AFEPQ, BFDPQ, and CDEPQ have a common point R. It is proved that when P, Q are the midpoints of the diagonals BE, CF respectively, then R lies on the third diagonal AD. Article 164 is The Transversal of a Quadrilateral . A transversal LMNP intersects a convex quadrilateral ABCD with L on AB etc. The harmonic conjugates of L, M, N, P are L', M', N', P'. It is proved that L'P' and M'N' intersect on the diagonal BD and that L'M' intersects N'P' on the diagonal AC. Article 165 is A description of properties of Pascal's hexagon . A description is given of properties of Pascal’s hexagon, illustrating how 6 Pascal lines fall into two sets of 3 concurrent lines, defining 2 Steiner points. Article 166 is Analytic treatment of a Romanian problem. . A triangle ABC is given and a conic that cuts each of its sides in two real points. The tangents at these points are drawn providing a hexagon of tangents circumscribing the conic, whose vertices are labelled DF'ED'FE as in Fig.1. It is shown that AD, BE, CF Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings are concurrent as are AD', BE', CF' and DD', EE', FF'. The positions of the resulting perspectives P, P' and Q are located. In the Romanian question the conic was the nine-point circle and then P lies on the hyperbola ABCKOi, where K is the symmedian point and Oi is the isotomic conjugate of the circumcentre O. The isotomic conjugate of P then lies on the diameter of the circumcircle joining the Steiner point and the Tarry point. It is also shown that in the general case AD', EF', FE' are concurrent as are BE', DF', FD' and CF', DE', ED'. Areal co-ordinates are used throughout with ABC the triangle of reference. Article 167 is Basic Properties of a Quadrangle possessing an Incircle . Properties of a quadrangle possessing an incircle. Article 168 is Properties of a pair of Diametrically Opposite Triangles . Starting from a triangle ABC and its circumcircle centre O, triangle DEF is such that AD, BE, CF pass through O. Tangents at the six points and the nine lines joining their vertices create a configuration that is explored in this article. Article 169 is Two In-Perspective Triangles inscribed in a Conic . Triangles ABC and DEF are inscribed in a conic and are in perspective through a point O. Polar lines LMN and PQR are drawn and six points of concurrency are shown to define a number of important collinearities. Two conics also emerge. Article 170 is The Brocard Conics . Triangles ABC and DEF are inscribed in a circle and AD, BE, CF are concurrent at the symmedian point K. Sides of the two triangles are extended and the tangents at the vertices are drawn. The polar lines of the two triangles coincide and points of intersection of the sides and tangents are shown to produce in addition four six-point conics all with their centres on the Brocard axis OK, where O is the circumcentre of ABC. Article 171 is A Conic through the feet of two Cevians may lead to a second Conic. Points P and Q are two Cevian points so that the feet U, V, W, U, V', W' carry a conic then the chords UV, WV', W'U' intersect as shown if and only if ABCDEF is a conic. Other properties (not proved in this article) hold as may be inferred from Fig. 1. Article 172 is The Inner Circle of a Cyclic Quadrilateral. Given a cyclic quadrilateral ABCD with midpoints of sides P, Q, R, S and midpoints of diagonals U, V, then it is well-known that a rectangular hyperbola passes through P, Q, R, S, U, V as well as O, the centre and H, J the other two diagonal points. We find the equation of this hyperbola and that of the inner circle OUVT, centre G. The midpoint W of UV is also the midpoint of PR and of QS. Article 173 is The Auxiliary Circles of a Cyclic Quadrilateral. First the cyclic quadrilateral ABCD is drawn. The midpoints of AB, BC, CD, DA, AC, BD are labelled P, Q, R, S, U, V, the centre of ABCD is the point O and the diagonal points are T = AC∧CD, H = AB∧DC and K = AD∧BC. (These are all finite points.) The 10 point rectangular hyperbola P, Q, R, S, O, U, V, T, K, H is shown. We establish the existence of the three midpoint circles OUVT, OPRH, OQSK and also the Semi-Diagonal point Circle LMNXYZ, where L, M, N are the midpoints of OT, OH, OK and X = OK∧HT, Y = OH∧KT and Z = OT∧H.
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Christopher Bradley's Geometry Writings Article 174 is A Cevian Circle leads to Perspective Triangles. A circle is drawn through U, V, W the feet of a Cevian point T. Three other points are created on the sides of triangle ABC enabling a second triangle DEF to be drawn. It is shown that triangles ABC and DEF are in perspective. Article 175 is When two Triangles intersect in Conics and in Circles. If P is internal to a triangle and points D, E, F are defined on AP, BP, CP respectively so that AD/AP = BE/BP = CF/CP then the intersections of triangles ABC and DEF lie on a conic. If P is the symmedian point K then the conic is a circle. Article 176 is Orthologic Triangles. Triangles ABC and DEF are orthologic if the perpendiculars from the vertex of one of them to the corresponding sides of the other are concurrent. The property is symmetric but not transitive. Article 177 is Special Pascal Lines. By considering the hexagon as two triangles in perspective and labelling their vertices as well as those of the defining conic, it is shown that two of the Pascal lines pass through points that are points of concurrence of three lines. Article 178 is More Circles centred on the Brocard Axis. Circles through B and C and the Brocard points lead to two circles centred on the Brocard axis and several concurrences. Article 179 is The Second Brocard Triangle. Given the Brocard points Ω, Ω' in a triangle the point of concurrence of the intersection of the line AK with the 7-point circle and circles AΩC, BOC, AΩ'B (D in the above diagram) and two similar points E, F found by cyclic change of A, B, C forms what is known as the Second Brocard Triangle (David Monk, private communication). An analysis is the content of this document. Article 180 is Circles through the Brocard points and the Circumcentre. With Brocard points Ω and Ω', circles are drawn through AΩΩ', BOW', COW and they meet at a point D on the circumcircle. Points E, F are similarly defined. It is proved that the lines AD, BE, CD are all parallel to ΩΩ'. Further, if circle AΩΩ' meets AB at X and AC at Y then XY is also parallel to ΩΩ', as are two further lines defined in a similar fashion to XY by cyclic change. Article 181 is Circles through the Brocard points and the Symmedian Point. Circles BΩK, CΩK both pass through the same point D on BC. E and F are defined by cyclic change and lie on CA and AB. DEF turn out to lie on a line parallel to the tangent at K to the 7-point circle (and hence perpendicular to the Brocard axis). Article 182 is Problems requiring Proof. Here are four problems. In fact the solution to problem 4 is known, but the other problems may well be open. Article 183 is Conics and Triangles in Perspective. It is shown how a pair of triangles in perspective lead to a conic and conversely. Article 184 is Cevian Perspectivity. Triangle ABC and PQR with Cevian points D and S respectively are said to be in Cevian perspective if the points 11 = AD∧PS, 22 = BD∧QS, 33 = CD∧RS are collinear. If Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings points are labelled as follows: 12 = AD∧QS,13 = AD∧RS, 21 = BD∧PS, 23 = BD∧RS, 31= CD∧PS, 32 = CD∧QS, then the following results hold: (i) triangles 13 23 21 and 31 32 12 are in (ordinary) perspective through a vertex X, (ii) the points 13 23 21 31 32 12 are co-conic and (iii) X lies on the line SD. Article 185 is The Four Conic Theorem. Given two sets of four lines which intersect one on one in four points in a straight line, then the twelve remaining points carry four conics in which through point two conics pass. Article 186 is 6 Points and 4 perspectives If 6 points provide 1 pair of triangles in perspective then there are 3 more pairs of perspective triangles, with the same vertex, in the same diagram. Their Desargues' axes create the sides of a triangle and a related transversal. (It is assumed the original pair of triangles are not in triple perspective.) Article 187 is A Perspective in a pair of Cyclic Quadrilaterals. When two cyclic quadrilaterals in the same circle are in perspective then an axis of perspective joining the intersections of all six pairs of corresponding sides exists. Article 188 is Are three Triangles ever in Mutual Perspective? The question in the title has the answer `Yes', for example three triangles in the Brocard porism are in fact in triple reverse perspective with each other. However, we study in this very short article the condition that three triangles must obey if they are to be in mutual perspective. (Perspectives, of course, are symmetric, but not generally associative.) Article 189 is The Midpoint Conic. It is familiar that the feet of a pair of Cevians are co-conic. We prove here that their midpoints are also co-conic. Article 190 is The Brocard Circles' Twin Conic The construction of the Triplicate Ratio Circle and the Brocard Circle is replicated but starting with the circumcentre O rather than the symmedian point K. It is found that the Brocard Circle is replaced by a conic (the twin conic) passing through O and K and having centre the midpoint of OK. This result leads to the concept of a conjugation whose properties we briefly outline. Article 191 is Three Circles and their Centres Given a cyclic quadrilateral and a point P not on a side, lines may be drawn through P parallel to its sides each line meeting the adjacent sides in a point. The eight points created in this way determine two circles and the three circle centres and P form a parallelogram. Article 192 is Generalization of the Triplicate Ratio Circle Given a triangle ABC a point P not on its sides is selected and lines parallel to AB and AC are drawn through P to meet the sides in four distinct points. The position of P is determined when these four points are concyclic. Article 193 is A Two Circle Problem Involving the GK axis Through a point X lines are drawn parallel to the sides of ABC. From the resulting six points two circles are drawn and the condition on X for their common chord to pass through X is determined, the result being that X must lie on the GK axis.
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Christopher Bradley's Geometry Writings Article 194 is The KG Nine-Point Conic The properties of a Nine-Point Conic are exhibited using Cevian point K (as well as the centroid G). The symmedian point Km of the medial triangle (X141) is shown to lie on the KG axis, the midpoint of KKm being the centre X of the nine-point conic. Article 195 is The Orthocentre Circle of a Cyclic Quadrilateral With ABCD a cyclic quadrilateral perpendiculars are drawn from A to BC and to CD meeting those sides at L and N respectively and perpendiculars are drawn from B to DA and to CD meeting those sides at Z and X respectively. Further the perpendiculars from C to AB and DA are drawn meeting those sides at M and P respectively and finally perpendiculars are drawn from D to AB and BC meeting those sides at Y and W respectively. Point A' = BZ^DY, point B' = CM^AL, point C' = BX^DW and point D' = AN^CP. It is found that A', B', C', D' are concyclic and The circle we term the orthocentre circle as it mirrors the construction of the orthocentre in a triangle. Nine other circles arise out of the figure and their properties are investigated. Article 196 is The Two Central Lines in a Cyclic Quadrilateral There are five main central points in a cyclic quadrilateral: the circumcentre O, the centroid F, the centre of mass G, the intersection of the diagonals E and the intersection of the lines joining the midpoints of opposite sides T. It is proved in this article that GTE is a straight line and that OTF is a straight line. Furthermore ET = 3TG and OT = 3TF and so GF is parallel to OE. (The anticentre also lies on OTF.) Article 197 is When Two Pairs of Diagonals are Concurrent If ABCD is a cyclic quadrilateral and the tangents at A, B, C, D form a quadrilateral PQRS, the diagonals AC, BD and PR, QS all pass through a given point. Here we identify this point. Article 198 is The Two Cevians' Perspective In a triangle Cevians are drawn through P and Q. The points L, M, N, R, S, T are defined as follows: L = BQ^CP, M = BQ^AP, N = AQ^CP, R = CQ^BP, S = AQ^BP, T = CQ^AP. It is proved that triangles LMN and RST are in perspective. Article 199 is Three Circles with collinear Centres With ABCD a cyclic quadrilateral centre O and midpoints of sides PQRS, lines PO, QO, RO, SO meet the opposite sides at T, U, V, W respectively. Circles SUQW and TVPR have centres F and E respectively. It is proved that FOE is a straight line. Some other obvious results are mentioned. Article 200 is Problem on Quadrilateral with an Incircle Given a quadrilateral with an incircle there are vertices A, B, C, D and points of contact P, Q, R, S. Lines AQ, AR, CP, CS are drawn intersecting in K and M. Lines BS, BR, DP, DQ are drawn intersecting in N and L. It is proved that K, M lie on BD and N, L lie on AC. Article 201 is All about a Cyclic Quadrilateral and its cousin Given a Cyclic Quadrilateral ABCD with circumcentre O and diagonals AC, BD, EF then the midpoints of the diagonals P, Q, R respectively are well known to be collinear. The points U, V, W are the intersections of the diagonals. It is first shown that circle UPQ passes through O and that if the line RU meets this circle at S, then the following circles all pass through S. These circles are VRP, WRQ, UVW, EFO, ACR, BDR. If the circles WVA, WVB, WVC, WVD meet the circle ABCD at points A', B', C', D' then RAA', RBB', RCC', RDD' are all straight lines. It is not unreasonable to say that S is the most important point in the figure.
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Christopher Bradley's Geometry Writings Article 202 is The Midpoint Theorem Let ABCD be a quadrangle, with L, M, N the midpoints of AB, BC, CA and U, V, W the midpoints of AD, BD, CD, then the conics AULMN, BVLMN, CWLMN have a fourth point in common. (D must not lie on the medians of ABC.) Article 203 is Another Circle with Centre on the Brocard axis In triangle ABC with symmedian point K, circumcentre O, circles BKC, CKA, AKB meet the sides at six further points, two on each side. It is shown these points lie on a circle with centre at a point O' on the Brocard axis such that OK = 2KO'. Article 204 is Between the Triplicate Ratio Circle and the Circumcircle When UVW is a triangle homothetic (and similar) to triangle ABC through K then the six points where UV, VW, WU intersect the sides of ABC lie on a circle. When UVW becomes a point at K one gets the Triplicate Ratio Circle and when U, V, W reach A, B, C one gets the Circumcircle. The centres O' of these circles move along the Brocard axis. Article 205 is The Brocard Lines KΩ and KΩ' Circles AΩΩ', BΩΩ', CΩΩ' meet the sides of ABC also in six further points which lie on two lines KΩ, KΩ' (which we call the Brocard lines). These six points also lie on four circles, each of the six points lying on two of the circles. Article 206 is Dividing OH into Five or Seven Equal Parts By dividing each side of a triangle into quarters and then drawing the resulting nine Cevian lines, six points are created internally through which two circles are drawn. Their centres lie on the Euler line and divide OH in the ratios 1:4 and 4:3. Article 207 is The Median Conic Given a triangle ABC and its centroid G, the midpoints of AG, BG, CG are denoted by U, V, W. Conics BCUVW, CAUVW, ABUVW are drawn and meet the sides again at point D, D', E, E', F, F'. It is proved that D, D', E, E', F, F' lie on a conic. Article 208 is A Pair of Isogonal Conjugates produce a Conic through 6 Points If P and Q are an isogonal conjugate pair then circles APQ, BPQ, CPQ intersect the sides of triangle ABC in six points and it is found that a conic passes through them. The construction only produces a conic when P and Q are isogonal conjugates. When P and Q are the Brocard points then the conic degenerates into a pair of lines. Article 209 is Three Concentric Circles Starting with a cyclic quadrilateral ABCD with centre O, the midpoints A', B', C', D' of OA, OB, OC, OD respectively form a second cyclic quadrilateral with half the size. It is shown that points B, B', C', C lie on a circle and so do C, C', D', D etc. The centre of BB'C'C is labelled bc, and that of CC'D'D is labelled cd etc. Five pairs of these circles are immediately seen to be coaxal and analysis is given for one such pair. Article 210 is 12 Points, 8 six-point Conics, 4 Conics through each Point Given a triangle ABC from the midpoints of each side perpendiculars are drawn to the other two sides. These lines are labelled Lines 1 to 6 as in the Figure. Their intersections create 12 finite points, so that, for example 35 is the intersection of Lines 3 and 5. It is found that 8 six-point conics can be drawn through these 12 points, with 4 of the conics through each point. The nine-point circle is, of course, one of the conics. In a later article some properties of these conics are established.
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Christopher Bradley's Geometry Writings Article 211 is A cyclic quadrilateral and its midpoint circles A cyclic quadrilateral its midpoints and The Six Midpoint Circle Article 212 is Quarter size Circles at Triangle Vertices A cyclic quadrilateral ABCD with AC perpendicular to BD is such that the tangents at A, B, C and D form a cyclic quadrilateral. This configuration is described analytically. If AC^BD = X and the centres of the two circles are O and Y, then it is proved that YOX is a line. Article 213 is When an Incircle produces a Circumcircle A cyclic quadrilateral ABCD with AC perpendicular to BD is such that the tangents at A, B, C and D form a cyclic quadrilateral. This configuration is described analytically. If AC^BD = X and the centres of the two circles are O and Y, then it is proved that YOX is a line. Article 214 is The Super-Cevian Triangles, their Conic and Two Perspectives In the above Figure, ABC is a triangle, P is a Cevian point and D, E, F the feet of the Cevians on BC, CA, AB respectively. The triangle DEF is the Cevian triangle. A triangle XYZ is constructed by drawing through A a line parallel to EF and through B and C lines parallel to FD and DE respectively. Also a triangle UVW is constructed by drawing lines parallel to EF, FD, DE through D, E, F respectively. These two triangles we term as SuperCevian triangles. We then prove that ABCUVW is a conic. Q is the perspector of triangles UVW and XYZ and R is the perspector of triangles ABC and XYZ. These results are established in this article, using areal co-ordinates with ABC as triangle of reference. Article 215 is Multiplication of Points using Barycentric Co-ordinates If you have points with barycentric co-ordinates (f, g, h), (u, v, w) then the rule of multiplication is that the result has co-ordinates (fu, gv, hw). It is shown in this article how to perform this multiplication using a geometric construction . The method is as follows: First draw the circumconic fyz + gzx + hxy = 0. Next take the isotomic conjugation of (u, v, w) to get the point (1/u, 1/v, 1/w). To conjugate this point finally use the circumconic to perform the second conjugation taking (1/u, 1/v, 1/w) to the product point (f/(1/u), g/(1/v), h/(1/v)) = (fu, gv, hw). This is evidently a commutative product and so the product may also be obtained using the conic ux + vy + wz = 0 and operating the two conjugations on the point (f, g, h). Article 216 is A Point, two Triangles and Two Conics A triangle ABC is given along with its circumcircle Γ and any internal point P. Lines AP, BP, CP are extended to meet Γ at U, V, W respectively. Triangle UVW is drawn and their sides intersect internally in six points. BC meets AU at a, with b and c similarly defined. AU meets side VW at u with v and w similarly defined. It is shown in this article that a conic passes through the first six points and that a conic also passes through a, b, c, u, v, w. Article 217 is An interesting Perspective in the Anticomplementary Triangle A triangle ABC and its anticomplementary triangle A' B'C' are drawn and a point P is selected through which Cevians A'P, B'P, C'P are drawn meeting the sides of ABC in points L', M', N' and the sides of A', B', C' in points L, M, N. Circles LMN and L'M'N' are drawn meeting the sides of A'B'C' and ABC respectively in points U', V', W' and U, V, W. Several perspectives are created, but the most interesting is that of triangles A'B'C' and UVW with perspector Q.
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Christopher Bradley's Geometry Writings Article 218 is Properties of the Triangle of Excentres In the triangle of excentres the orthocentre is the incentre I of ABC, the Symmedian point K+ is the Mittenpunkt of ABCand the nine-point centre is the circumcentre O of ABC. The circumcentre O+ lies on the line OI and is such that O+O = OI. These properties are proved as also the fact that KI passes through K+. Article 219 is The Twelve point Circle In the triangle of excentres the orthocentre is the incentre I of ABC, the Symmedian point K+ is the Mittenpunkt of ABCand the nine-point centre is the circumcentre O of ABC. The circumcentre O+ lies on the line OI and is such that O+ O = OI. These properties are proved as also the fact that KI passes through K+. Wow! Article 220 is A Conical Hexagon with Main Diagonals Concurrent A hexagon ABCDEF is inscribed in a conic with diagonals AD, BE, CF concurrent at a point P. It is shown that this property is replicated and that both conics have the same polar with respect to P. This polar is one of the Pascal lines. Article 221 is Four Special Conical Hexagons all with same Polar Given a hexagon ABCDEF inscribed in a conic with AD, BE, CF concurrent at P, the tangents tA, tB, tC, tD, tE, tF are drawn. Points jk = tj^tk are determined and it is shown that 12, 23, 34, 45, 56, 61 lie on a conic. Lines 61 12, 34 45 meet at point 1, lines 12 23, 45 56 meet at 2 and lines 23 34, 56 61 meet at 3. Points 1, 2, 3 are collinear and it is shown that 123 is the polar of P with respect to both conics. Article 222 is The Missing Point on the Euler Line Given a triangle ABC, circumcentre O, orthocentre H, let the midpoints of sides be L, M, N and suppose AO, BO, CO meet the sides BC, CA, AB respectively at U, V, W then the centre of the ellipse LMNUVW is a point S on the Euler line such that OH = 4OS. There appears to be no reference to this point in the literature, and so we call it The Missing Point. Article 223 is When a non-regular Cyclic Pentagon leads to Another It can be arranged that a non-regular cyclic pentagon ABCDE, by special adjustment of D and E, produces a second cyclic pentagon A'B 'C'D'E', where A' = BD^CE, B' = AC^DE, C'= AB^DE, D'= AB^CE and E' = AC^BD. Article 224 is The Sixty Pascal Poles When six points lie on a conic then sixty Pascal lines may be constructed. In this paper it is shown how the poles of sixty these lines may be constructed without drawing tangents. Their properties are, of course, the duals of those of the Pascal lines. Article 225 is A Typical Pascal line Drawing ABCDEF is a cyclic hexagon (though the results remain true if the circle is replaced by a conic). A'B'C' is a Pascal line, where A' = BC^AF, B' = CD^AF and C' = DE^AB. We denote this by (AEC, DBF), where care must be taken with the order of letters. Points X, Y, Z are given by X = AC^DF, Y = AE^DB, Z = EC^BF. Points X', Y', Z' are given by X' = AD^CF, Y' = AD^EB, Z' = CF^EB. In Article 225 it is shown that XX', YY', ZZ' are concurrent at a point P which is the pole of A'B'C' with respect to the circle. Article 226 is Special Conical Hexagons Hexagon ABCDEF inscribed in a Conic is said to be special if AD, BE, CF are concurrent at a point P. When this happens hexagons formed by taking the intersections of neighbouring tangents are proved also to be special and conical with the same point P.
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Christopher Bradley's Geometry Writings Article 227 is The Remarkable Eight-Point Conic Two conics intersect at A, B, C, D and AC meets BD at P. A line through P is drawn meeting one of the conics at F, G and the other at E, H. Tangents are drawn at F, G, E, H. Tangents at F and G meet at Q and tangents at E and H meet at R. Tangents at H and G meet at V, tangents at H and F meet at S, tangents at E and F meet at T and tangents at E and G meet at U. A conic now passes through the 8 points A, B, C, D, S, T, U, V and the line QR is the polar of P with respect to all three conics. The general case is technically too difficult to establish, so here we provide a numerical case. Article 228 is How 2 Conics through 4 Points generate 6 more such Conics Two Conics through 4 point A, B, C, D are drawn (in the figure the light blue and the orange). Their tangents to the first (the orange) at A, B, C, D are labelled 1, 2, 3, 4 and the tangents to the second (the light blue) are labelled 5, 6, 7, 8. Intersections of these tangents are labelled 12, 13, 14, 23, 24, 34 and 56, 57, 58, 67, 68, 78. It is now found that six 6 point conics may be drawn. Article 229 is When a Conical Quadrilateral produces two 6-point Conics If a quadrilateral ABCD is drawn in a Conic and the tangents are drawn at A, B. C. D then the sides of ABCD and the intersections of the tangents produce two six point conics and a polar line. Article 230 is A Quadrilateral and the Conics and Points arising ABCD is an arbitrary quadrilateral inscribed in a general conic. The tangents at its vertices form a second quadrilateral PQRS. There emerges the polar line FGTU and 8 points of intersection of ABCD and the tangential quadrilateral PQRS. Many conics result, but two of them in particular meet on the diagonal PR at points X, Y in the figure. X and Y appear to have properties as important as any of the 8 vertices of the quadrilaterals and the four points on the polar line. Not by any means are all the conics that can be drawn featured in this article and a second article is envisaged. What determines why the diagonals PR and QS have such different properties are (as far as we are concerned) open and challenging questions. Article 231 is A Quadrilateral and resulting Conics Part 2 A Special Case showing the role of the New Points X and Y In Article 230 we investigated properties of a configuration of a quadrilateral ABCD inscribed in an ellipse and the common points of tangents to the ellipse at the vertices A, B, C, D. It turns out that there are two remarkable points X and Y through which many lines may be drawn and also through which many conics pass. The analysis of Part 1 was too complicated to illustrate the results, so in this Article we give a numerical presentation that illustrates the properties of the point X and Y, for which the results all hold in the general case. It seems unlikely the general case can be treated analytically, but possibly a pure approach would succeed as the points are members of a Desargues' involution. Figure 1 above illustrates the line properties of X and Y and Figure 2 below illustrates their Conical properties. Article 232 is The Coconic Hexagon with Main Diagonals Concurrent ABCDEF is a hexagon inscribed in a conic and is such that its main diagonals AD, BE, CF are concurrent at a point G. Sides AB, BC, … are labelled 1, 2,… . Points such as 14 are then the intersection of AB and DE. Tangents are drawn at A, B, …, so that the tangents at A and B meet at P. Point Q is the intersectionof the tangents at B and C and so on. The tangents at A, B, … are also labelled tA, tB, … and points such as ac are thus the intersections of tA and tC. It is proved that PQRSTU is a conic and that the six points ac, bf, ae, df, ce, bd lie on a conic. It is also proved that 13, 26, 15, 46, 35, 24 lie on a conic.
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Christopher Bradley's Geometry Writings Article 233 is The Conic and the Lines that are Created (Part 1) A Conic is defined by 5 points, no three of which are collinear, Pairs of lines joining these points and the tangents to the conic at its defining points create 10 lines and these lines generate 15 other key lines and many subsidiary ones. Article 234 is A Conic generates five 8-point Conics (Part 2) A conic ABCDE with sides 1 = AB, 2 = BC etc. and its tangents a, b, … at the vertices AB, … produce 20 intersections through which five 8-point Conics may be drawn. Article 235 is The Truth and the whole Truth about a Quadrilateral A quadrilateral is inscribed in a conic and its sides and diagonals and the tangents at its vertices are drawn. A study is made of the lines and conics that pass through the points of intersection. In all a total of eighteen conics emerge three of which pass through eight points and fifteen through six points. Article 236 is A Nice Conic and a Nasty Circle A triangle ABC and its circumcircle S are given. The line through B parallel to CA and the line through C parallel to AB meet at L. Points M, N are similarly defined by cyclic change. Tangents to S at B and C meet at U, with V, W defined by cyclic change. It is proved that LMNUVW is a conic and the centre of circle UVW is shown to lie on the Euler line of ABC. Article 237 is How to Construct the Ex-points of a given Point It is shown how knowing the Ex-points of one point it is possible to construct the Ex-points of other points. Article 238 is When Two Quadrilaterals are in Complete Perspective If two quadrilaterals ABCD and PQRS are such that AP, BQ, CR, DS are concurrent at X then S may be moved into one and only one new position on DX so that the four Desargues' axes coincide. Article 239 is More on Complete perspective It is shown that if two quadrilaterals ABCD and PQRS are such that AP, BQ, CR, DS are concurrent at a point X (and with the notation that AB^PQ = ab etc.) then if ab, bc, cd are collinear the quadrilaterals are in complete perspective (see Article 238). Article 240 is Pentagons in Complete Perspective If ABCDE and PQRST are pentagons such that AP, BQ, CR, DS, ET are concurrent at a point X. Using the notation AB^PQ = ab etc. it is proved that if ab, bc, cd, de are collinear then the pentagons are in complete perspective, that is the ten Desargues’ axes merge into the one line ab bc. Article 241 is Quadrilaterals in Perspective inscribed in a Conic If ABCD is a quadrilateral inscribed in a conic S and P is any point not on S and AP, BP, CP, DP meet S again in points A', B', C', D' then the quadrilaterals ABCD and A'B'C'D' are in complete perspective. In particular if AC^BD = E and A'C'^B'D' = E' then E, E', P are collinear.. Article 242 is Perspective Quadrilaterals Two quadrilaterals ABCD and A'B'C'D' are such that AA', BB', CC', DD' are concurrent at a point X. It is found that a single condition is such that the four Desargues’ axes of perspective coincide. It is also the case that when this happens the diagonal point triangles EFG and E'F'G' are also in perspective with vertex of perspective X and the same Desargues’ axis of perspective. See also Article 238.
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Christopher Bradley's Geometry Writings Article 243 is Some properties of a Special Hexagon The hexagon ABCDEF is such that AD, BE, CF are concurrent at a point U. Given the quadrilaterals ABDE, ACDF, BCEF the diagonal points (other than U) are labelled V, W, V', W', V'', W''. It is proved that these six points lie on four lines with three points on each line and such that each of the six points lies on two of the lines. Article 244 is When two Triangles in Perspective create Conics When triangles ABC and DEF are in perspective through a point I a conic does not normally pass through A, B, C, D, E, and F. It is proved here if the position of F is moved along CI a position may always be found for F so that a conic passes through all six points. When this is so other conics may be drawn through suitably created points. Article 245 is Some Simple Algebraic Results Part 1 The conic passing through (1, 0, 0), (0, 1, 0), (0, 0, 1), (u, v, w), (p, q, r). Part 2 Some results about the conic y2 = zx. Part 3 The conic yz + zx + xy = 0 Part 4 A conic containing 3 pairs of points in perspective Article 246 is Pascal points An analysis of a Pascal point is given, using the conic y2 = zx and general points. Article 247 is The G Conic Through the centroid G lines parallel to the sides are drawn intersecting the sides in six points. A conic passes through these six points, which we call the G Conic. Its centre is G and we obtain its equation. Article 248 is The H Conic Lines are drawn parallel to the sides through the orthocentre H of a triangle ABC meeting the sides in six points. The conic through these six points we term the H conic. We obtain its equation and the co-ordinates of its centre X. It is shown that X, H and K, the symmedian point, are collinear. Article 249 is Twin Conics In a triangle ABC if lines through the symmedian point K are drawn parallel to the sides to meet opposite sides in 6 points, then it is well known that the conic through these six points is a circle called The Triplicate Ratio Circle. If K is exchanged for another point Q and the same construction is carried out then a conic passes through the 6 points which we call the Q conic. If X is the centre of this conic then we show there is a second point Q' also having X as centre and arising from the same construction. If Q is at the centroid G then Q' is also at G. We show that the Triplicate Ratio Circle is twinned to the O conic and we determine the twin of the H conic. The general point Q is first treated and the equation of the Q conic is obtained as also its centre X. The twin conic, the Q' conic, is also obtained. Article 250 is Cevian Circles The feet of a set of Cevians through a point S are three points through which a circle may be drawn. The other three points where the circle cuts the sides are also the feet of a set of Cevians through a point T. The co-ordinates of T are obtained in terms of those of S. When S is the centroid G then T is the orthocentre. There do not appear to be any other important circles deserving the title Cevian Circles. Article 251 is The Four Brocard Triangles The six circles used to construct the Brocard points are analysed to produce the First and Second Brocard triangles situated on the circumference of the Brocard Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Christopher Bradley's Geometry Writings circle and similar constructions produce the Third and Fourth Brocard triangles which turn out to be the Isogonal Conjugates of the first two. Article 252 is Odds and Ends Part 1: Equation of Circle from radius and centre
nd
Part 2: Change of Brocard points and the construction of the 2
and 4
th
B-Triangles
Article 253 is Triangles in Perspective – the Whole Truth Triangles ABC and 123 are in perspective if A1, B2, C3 are concurrent at a point X commonly called the perspector. It then follows that the points AB^12, BC^23, CA^31 are collinear. The line is called the Desargues’ axis of perspective and nowadays is often called the perspectrix. That is all that is taught to elementary classes. However, it is in fact far from the end of the matter. If we define the cross-over triangle to have vertices B1^A2, C2^B3, A3^C1 then we show that B1^A2, C2^B3 and CA^31 are collinear as are B1^A2, A3^C1and BC^23 and A3^C1, C2^B3 and AB^12. In other words there is not just an axis of perspective but a complete quadrilateral of perspective. But even that is not all. We prove also that the remaining six points A2^B3, B1^C2, B3^C1, C2^A3, A3^B1, C1^A2 lie on a conic. This last result I have not yet located in past literature. Article 254 is Circles through the midpoints of sides of a Cyclic Quadrilateral Let ABCD be a cyclic quadrilateral with diameter AC, with centre O and with external diagonal points Q and R and let S, T be the midpoints of AQ, AR and U, V the midpoints of BC, CD. Then circles QUVR and BOSTD exist and their line of centres passes through the midpoint of AO. Article 255 is Five Conics and a Polar line Two triangles inscribed in a triangle are also in perspective. Five Conics are now created and a polar line which is also two Pascal lines. Article 256 is Two Quadrangles in Perspective in a Conic Two Quadrangles in Perspective in a Conic are drawn. We prove their six diagonal points lie on a conic. Article 257 is Quadrangles in Perspective Part II The three lines Two quadrangles are drawn in a conic and are in perspective. Three lines emerge with six points of intersection on each line. They are like Pascal lines but appropriate to Quadrilaterals in perspective and not to pairs of triangles. It is difficult to be sure but they may be a new discovery. Article 258 is The Miquel Cyclic Quadrilateral Let ABCD be a cyclic quadrilateral with points L, M, N on the line segments AB, CD, DA respectively. Circles ANL and DMN meet at N and a point P (which we assume does not lie on AC or BD – otherwise adjust L, M, N). Circle BPL now meets BC at K. We now prove that circle CPM passes through K. We thus have a Miquel point P with four circles through it. The Miquel point is not automatic since the point K has to arise naturally. Circles ANL and CMK meet at P and another point R that we show lies on AC. Circles DMN and BLK meet at P and another point S that we prove lies on BD. We then show circle PRS passes through Q the intersection of AC and BD. Article 259 is Two Triangles and an Eight Point Conic Let ABC and TUV be two triangles with L, M, N the points BC^UV, CA^VT, AB^TU respectively Suppose adjustments are made so that L, M, N are the feet of Cevians of both triangles. Then we prove a conic passes through A, B, C, T, U, V and the two Cevian points. Romantics of Geometry (https://web.facebook.com/groups/parmenides52/)
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Article 1 The Story of Hagge and Speckman Christopher J Bradley
1. Introduction In 1905 Speckman [1] wrote a paper on indirectly similar triangles in perspective and in 1907 Hagge [2] wrote another paper on the construction of circles that always pass through the orthocentre of a given triangle, which have become known as Hagge circles. It is evident that Hagge was unaware of Speckman’s work, and indeed it appears that the contents of Speckman’s paper, possibly because it is written in Dutch, seem to be little known. Hagge’s work, on the other hand has been developed, notably by Peiser [3], and more lately by Bradley and Smith [4], [5]. Both the 1905 and 1907 papers are ground breaking papers in classical geometry, and the work of Speckman, in particular, deserves to be better known. The interesting thing is that the two pieces of work are related in a manner that at first sight is not at all obvious. However, if Hagge had read Speckman’s paper it is almost certain that he would have understood its relevance and we would not have had to wait so long to see the developments of Hagge’s work that are a consequence. For reasons that will become clear it is best to describe Hagge’s work first and then to summarise Speckman’s work, showing how it may be applied, step by step, to Hagge’s configuration. 2. Hagge circles For the construction of a Hagge circle we refer to Fig.1. Given a triangle ABC and a point P not on a side or the extension of a side and not on the circumcircle, the Hagge circle of P with respect to ABC is defined as follows: draw AP, BP, CP to meet at D, E, F respectively; now reflect D, E, F in the sides BC, CA, AB to obtain the points U, V, W. The Hagge circle (P) is now defined to be the circle UVW. In his paper [2] Hagge proved (i) that (P) always passes through H, the orthocentre of triangle ABC; (ii) that, if (P) intersects the altitudes AH, BH, CH at X, Y, Z respectively, then UPX, VPY, WPZ are straight lines; (iii) that the Hagge circle of G, the centroid of ABC, with respect to the medial triangle (the triangle joining the midpoints of the sides) is the Brocard circle on OK as diameter, where K is the symmedian point of ABC, O being the circumcentre of ABC and thus the orthocentre of the medial triangle. At the time, Brocard geometry was a subject of particular interest, which may account for why roughly half of Hagge’s paper is devoted to this application. 1
Peiser’s [3] main contribution was concerned with the centre Q of the circle (P). Using the algebra of complex numbers he was able to show that the image of Q under a halfturn about the nine-point centre T is the same point as the isogonal conjugate Pg of P with respect to triangle ABC. Since T is the midpoint of OH it follows that the figure PgHQO is a parallelogram. Another property not mentioned by earlier geometers is that the midpoints of AU, BV, CW all lie on the nine-point circle.
A
Pg Y H
F
T G U
O
W
E
P Q Z C
B V
X
D
Fig. 1
If P lies on the circumcircle the lines APg, BPg, CPg are parallel, so Pg lies on the line at infinity. The consequence is that the Hagge circle degenerates into a line through H, which being the line through the reflections of P in the sides BC, CA, AB is parallel to the Wallace-Simson line of P and is sometimes called the double Wallace-Simson line of P. When P is the incentre I the Hagge circle is the Fuhrmann circle and this circle was known about many years before Hagge’s publication. The Fuhrmann circle not only passes through H but also passes through Nagel’s point Na. An account of the Fuhrmann circle is given by Honsberger [5], who shows that HNa is one of its diameters. When P is the symmedian point K the Hagge circle is the orthocentroidal circle on GH as diameter. 2
In general there is a straightforward result that PgG meets the Hagge circle at the point at the other end of the diameter through H. See Fig. 1.1. In their first paper Bradley and Smith [4] give a synthetic proof of Peiser’s result and also show how a consistent definition can be given for a Hagge circle if P lies on a side of the triangle or at a vertex. In their second paper Bradley and Smith [5] provide another proof of Hagge’s result (ii). Now if you consider the quadrilateral HXYZ, since CZH and BHY are perpendicular to the sides, we have ZXY = ZHY = BAC. Similar angle relations show that triangles ABC and XYZ are similar, and since the ordering of the letters in the two triangles is opposite, these triangles are indirectly similar. The same applies to triangles UVW and DEF. Bradley and Smith [4] also give a synthetic proof of the important result that there is an indirect similarity carrying ABCDEFP to XYZUVWP in which P is the unique fixed point. In Fig. 2 we provide an illustration of the dilative reflection through P and its axis that provides this indirect similarity.
3
F A
V' X' V
Z' Z
E X
P W
Q
U Y U'
H C
B
D
Y'
Fig. 2
On reflection in the axis through P the figure ABCDEF is mapped to XYZUVW and then by dilation through the centre P by a factor XY/AB the figure XYZUVW is mapped to XYZUVW. Note that if the perpendicular axis is used instead, then a rotation of 180o is required in addition to the dilation and reflection. It is also significant, as we shall see in due course, that triangles ABC and XYZ are in perspective with vertex of perspective H, and that triangles XYZ and UVW are in perspective with vertex of perspective P. 4
W'
Further results proved by Bradley and Smith [5] are (i) that, if VW meets AH at U, WU meets BH at V and UV meets CH at W, then U, V, W, P are collinear, and (ii) that, if U1, V1, W1, X1, Y1, Z1 are the midpoints of DU, EV, FW, AX, BY, CZ, then U1, V1, W1, X1, Y1, Z1 lie on a conic, which they call the midpoint conic. As regards result (ii), it is in fact the case that if A, B, C, D, E, F are six points on a conic Σ and if X, Y, Z, U, V, W are their images on a conic Ψ under an indirect similarity then the midpoints of AX, BY, CZ, DU, EV, FW lie on a conic. In fact one can go further and say that the points that divide these six segments in any fixed ratio lie on a conic. 3. Speckman’s theorems on indirectly similar triangles in perspective Speckman announced proofs of his theorems in April 1903 at a Congress in ś-Gravenhage and they were published [1] in 1905. Their relevance to the work of Hagge lies in the fact that in the Hagge configuration, described in Section 2, triangles ABC and XYZ are indirectly similar and are in perspective with vertex of perspective H. If Hagge had read Speckman’s work, then it seems highly likely that he would have applied the theorems to obtain a sequence of additional results. And if Speckman had known about Hagge’s work it seems almost certain he would have referred to it subsequently. As far as we are aware the application of Speckman’s results to the Hagge configuration has never been pointed out and is a story waiting to be told. We tell this story by relating the more important of Speckman’s results paragraph by paragraph, pointing out in each case its significance in relation to Hagge’s configuration. Whilst it is true in an obvious sense that Speckman’s results are more general than Hagge’s (he deals with the case when the vertex of perspective is a general point and not the orthocentre and consequently derives more general results), it is possible to hold an alternative point of view that Hagge’s configuration is illustrative of all the properties one might need to know about pairs of indirectly similar triangles in perspective. For if you have an arbitrary pair of indirectly similar triangles in perspective, one of them is always related to the other by being directly similar to one of its partner’s Hagge triangles (defined to be the triangle XYZ inscribed in a Hagge circle). This is illustrated in Fig. 3. It shows two arbitrary triangles ABC and ABC that are indirectly similar by means of a reflection in the line l through P followed by dilation through P. They are chosen to be in perspective with vertex a point Q. Triangle XYZ is the Hagge triangle of P and triangle XYZ shows a dilation of XYZ through P with an appropriate scale factor. It can now be observed that triangles ABC and XYZ are images of one another under a rotation about P. If the point P is not known it may be obtained as the other point of intersection besides Q of the two rectangular hyperbolae ABCHQ and ABCHQ. See the first paragraph of Section 4.
5
Q
To Q
l
C' Y'
A Y
X'
X H
A'
P H'
Z
C
B B'
Z'
Fig. 3
4.
Paragraphs 1 to 9 of Speckman’s paper
We refer to Fig. 3 in which H and H are the orthocentres of triangles ABC and ABC, Q is the perspector and P is what Speckman calls the double-point of inverse similarity. He proves in the first paragraph that rectangular hyperbolas may be drawn passing through ABCHPQ and ABCHPQ and possessing asymptotes that are parallel. From now on we describe Speckman’s results primarily as they apply to the Hagge configuration. The more general results, if not stated, should be clear enough. In Fig. 4 the result of the first paragraph is illustrated when the second triangle is the Hagge triangle XYZ. Note that Q now coincides with H and the asymptotes are parallel to the axes through P, which Speckman calls the double lines of inverse similarity. Also he 6
shows in the second paragraph that these lines are parallel to the angle bisectors of corresponding lines of triangles ABC and XYZ. The reflection of XYZ in one of these axes is the triangle XYZ, which is a dilation of triangle ABC through P. Triangles ABC and XYZ are orthologic, as indeed are all pairs of indirectly similar triangles (a fact we prove later in this article) and the orthologic centres of the triangles are D and H, where the point D lies on both the circumcircle of ABC and the rectangular hyperbola ABCHP. Also, as shown in Bradley and Smith [4], the orthocentre h of triangle XYZ, P and D are collinear. The centres of the two rectangular hyperbolas are denoted by M and m. Also illustrated in Fig. 4 is the fact, proved in Speckman’s third paragraph, that J, j and P are collinear where j is the 180o rotation of j about m, and J, j are any pair of corresponding points on the two hyperbolae under the similarity.
J
Co
Bo D
j' Z
A X'
h
Ao
M m
X
P H
Y'
Z'
Y
j
B
C J'
Fig. 4
In paragraph four there is a construction, whereby triangle ABC is rotated by 180° about the centre M of the hyperbola ABCHP to produce a triangle A0B0C0 and it is proved that triangles XYZ and A0B0C0 are also inversely similar and in perspective, but now the 7
vertex of perspective is P, the double point of inverse similarity is H and the orthocentre of triangle A0B0C0 is D. In the more general case the interchange of the roles of the double point of inverse similarity P and the vertex of perspective Q arising from this construction is an intriguing result. In paragraph five there is proof of the result that corresponding points on the two hyperbolae together with the points obtained by half-turns about their respective centres M and m are concyclic. An arc of the circle JjjJ can be seen in Fig. 4. In paragraph seven there is a proof that in the general case (when the perspector is not at one of the orthocentres) the two orthocentres are collinear with the perspector. This result is trivial in the Hagge configuration, since Q then coincides with H. In paragraph nine it is proved that the lines connecting the orthological centre of one triangle to the orthocentre of the other passes through the double point of inverse similarity. When applied to the Hagge configuration this implies that the line hD passes through P, a result which is also proved in Bradley and Smith [5].
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M To M'
A
Z H
U
N' N B
V'
Y L W
U' V P
D
h C X
L' Fig. 5
5. Axes of perspective in the Hagge configuration In Fig. 5 we show the axis UVW defined at the end of Section 2 and also the Desargues axes LMN, LMN of perspective arising respectively from the perspectives of triangles ABC, XYZ and of triangles XYZ, UVW. Paragraph ten of contains the proof that the axis LMN is perpendicular to the line HD joining the orthologic centres.
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6. Paragraphs 11 to 18 of Speckman’s paper Fig. 6 illustrates the results of paragraphs 11-16. Most of the notation has previously been defined. Pg is the isogonal conjugate of P with respect to triangle ABC and pg is the isogonal conjugate of P with respect to triangle XYZ. In paragraph eleven it is noted that the line through A parallel to YZ, the line through B parallel to ZX and the line through C parallel to XY are known to meet and it is proved they do so at a point S on the circumcircle of ABC. Speckman calls this point the metapole of triangle ABC with respect to triangle XYZ. Such a point is now called a paralogic centre, so it turns out that we are dealing with a case in which the triangles are not only orthologic but also paralogic. (It is not always the case that orthologic triangles are also paralogic.)The point s is defined similarly. In the next two paragraphs there are proofs about the properties of the lines DS and Hs. They are diameters, respectively, of circles ABC and XYZ. It is proved that the line Hs is parallel to the isogonal conjugate of the rectangular hyperbola ABCHP, which is a line through Pg, and that DS is parallel to the isogonal conjugate of the rectangular hyperbola XYZHPh, which is a line through pg. Speckman calls the lines DS and Hs orthological diameters of the circles. A curious fact is that in the Hagge configuration the intersection R of the two orthological diameters lies on the rectangular hyperbola ABCHP. In paragraph fourteen the midpoints of the sides of ABC are denoted by Ma, Mb, Mc so that triangle MaMbMc is indirectly similar to triangle XYZ. It is shown that the perpendiculars from Ma on to YZ, Mb on to ZX and from Mc on to XY meet at a point W, which is the midpoint of HS. The point w is similarly defined, and is the midpoint of hs. Also HS/hs = AB/XY. In paragraph fifteen it is proved that the paralogic centre of triangle MaMbMc with respect to triangle XYZ is the midpoint of DH. This is the centre of the hyperbola ABCHPD. In paragraph seventeen it is shown that the Desargues axis of perspective LMN, see Fig. 5, bisects the line segment Ss joining the paralogic centres. In paragraph sixteen it is shown that if you reflect triangle XYZ in the Desargues axis LMN to get triangle XYZ, then triangle XYZ is in perspective with triangle ABC with vertex of perspective the point where DH meets the circumcircle ABC. In paragraph eighteen it is then noted that if the axis LMN meets the common chord of circles ABC and XYZ at E, then E is the radical centre of circles ABC, XYZ and XYZ. Hence the line through E perpendicular to OQ is the radical axis of the circumcircle and the Hagge circle. The construction of the radical axis of the circumcircle and the Hagge circle is a splendid flourish with which to end the paper.
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To E A Isogonal conjugate of ABCHP
D
Z
common chord of ABC and X'Y'Z'
my pg
s
M
h
w
X
X'
P
g
Q
mx
Mb
mz
Mc
O
G
H Y B
L
Z'
Pg Y'
C
Ma
S
To N
Isogonal conjugate of XYZHP
Fig. 6
11
7. Which of all these properties hold if there are two indirectly similar triangles that are not necessarily in perspective?
A
Z T
P
X
Y
C
B
Fig. 7
We first provide a construction. Take a triangle ABC, draw any circle (other than the circumcircle) and choose any point T lying on it. Draw lines through T parallel to the altitudes of triangle ABC to meet the circle again at points X, Y, Z. See Fig. 7. Simple angle arguments show that triangle XYZ is indirectly similar to triangle ABC. As T moves around this circle the triangle XYZ also rotates around the circle and CABRI shows there may be up to three positions of T for which triangles ABC and XYZ are also in perspective, but that is not a feature that interests us in this paragraph. The construction is such that the lines through X, Y, Z perpendicular to BC, CA, AB respectively are concurrent at T and hence triangles XYZ and ABC are not only indirectly similar, they are orthologic.
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Theorem 1.1 Every pair of indirectly similar triangles are orthologic and the orthologic centres lie on the circumcircles of the triangles. Proof Given a triangle XYZ that is indirectly similar to a triangle ABC, we can construct the circumcircle of XYZ and choose any point T on this circumcircle. We can then generate a triangle X'Y'Z' by drawing lines through T parallel to the altitudes of triangle ABC. Now all indirectly similar triangles inscribed in a given circle are congruent, so we can rotate T until triangles X'Y'Z' and XYZ coincide. The construction given at the outset of Section 7 now ensures that triangle XYZ is orthologic to ABC with orthologic centre at the final position of T. Triangles DEF and UVW are now defined as in the Hagge construction and they are similar as one is the image of the other under the indirect similarity. Paralogic centres S and s may be defined, even when there is no perspective. However, when there are no perspectives, there are no rectangular hyperbolas. But it is true in the more general case that the double lines of inverse symmetry are parallel to the angle bisectors of corresponding sides of triangle ABC and XYZ. When there is no perspective there is no Desargues axis of perspective LMN. However Cabri confirms that the axis U'V'W'P exists, as defined at the end of Section 2. The lines KS and Ts pass through the centres of their respective circles and so may justifiably be called orthologic diameters 8.
What happens when the centre of inverse symmetry lies at the orthocentre
When Speckman wrote his paper [1] on indirectly similar triangles in perspective he kept his account general and did not take account of particular cases. In other words his account assumes that the centre of inverse symmetry P, the perspector Q and the orthocentres of triangles ABC and XYZ are distinct points. The result is that he missed two special cases that are highly interesting. When the perspector Q coincides with H, the orthocentre of ABC, one gets triangles inscribed in circles through H and this gap was filled three years later by the ground breaking work of Hagge [2]. What has never been considered in the period from 1905 to the present day is what happens when the centre of inverse symmetry P lies at H. In fact it produces cases that are just as interesting as the Hagge circles that arise when Q lies at H. When P lies at H there are an infinite number of rectangular hyperbolas passing through A, B, C and H depending on the direction of the lines of inverse symmetry. And sure enough it turns out that if you choose any axis through H and then use it to form a triangle XYZ that is inversely similar to ABC, XYZ always turns out to be in perspective with ABC. The circumcircle of triangle XYZ obviously cannot be reduced by dilation through H to become a non-degenerate Hagge circle. In other words there are a whole set of pairs of indirectly similar and in perspective triangles and a whole set of circles, whose 13
properties have not been studied before. Of course, since Speckman just starts with a given set of points A, B, C, P and Q, his results are applicable when P lies at H, but he never considered what additional properties are true when either P or Q lies at H. And these additional properties are, as we shall see, quite substantial. Hagge circles are the outcome of what happens when Q lies at H. We now describe what happens when P lies at H. First, if you reflect ABC in any line through H, and use any enlargement factor k (k ≠ 0,1) for an enlargement (or reduction) centre H, then the resulting triangle XYZ is always similar to and in perspective with triangle ABC, with vertex of perspective at some point Q. Now let AH, BH, CH meet the circumcircle of ABC at D, E, F respectively. Through D draw a line parallel to AX to meet XP at U and define V, W similarly. Then it transpires that points U, V, W lie on the circumcircle of triangle XYZ and that triangle UVW is the image of triangle DEF under the indirect similarity. We know from work in Section 7, since XYZ is indirectly similar to ABC, there must exist points T' and T that are orthologic centres of the two triangles and that these points lie on the circles XYZ and ABC respectively and have the properties that the perpendiculars from T' on to BC, CA, AB pass through X, Y, Z respectively and the perpendiculars from T on to YZ, ZX, XY pass through A, B, C respectively. It also follows that T' lies on the rectangular hyperbola XYZHQ and T lies on the rectangular hyperbola ABCHQ. See Fig. 8 for illustration of all these properties. The earlier properties involving triangles XYZ, DEF and UVW require analytic proof and we now turn our attention to setting up this analysis.
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Q
A
Z T' U
E
J'
V
F Y
T
J
H
C
B
W
D
X
Fig. 8
Choosing co-ordinates In dealing with a triangle it is possible to choose three co-ordinates, such as the angular dispositions of the vertices and thereby make use of symmetry, or it is possible, by choosing the axes in some preferred directions, to choose just two co-ordinates. In the latter case symmetry is lost, but the expressions for the equations of lines and coordinates of points may be less complicated. In the present calculation which features an axis through H and dilation through H it is desirable for this point to be taken as origin. We therefore present a calculation in which the vertices have co-ordinates as follows: 15
A( – 2 – 2vw, 0), B(– 2vw, 2v), C(– 2vw, 2w). It may now be checked that H is the origin and the circumcentre O has co-ordinates (–1 – 3vw, v + w). The radius of the circumcircle R is given by R2 = (1 + v2)(1 + w2). The equations of BC, CA, AB are respectively x + 2vw = 0, y – wx = 2w(1 + vw), y – vx = 2v(1 + vw),
(8.1) (8.2) (8.3)
and the equations of AH, BH, CH are respectively y = 0, wy + x = 0, vy + x = 0.
(8.4) (8.5) (8.6)
The equation of the circumcircle Γ is x2 + y2 + 2(1 + 3vw)x – 2(v + w)y + 8vw(1 + vw) = 0.
(8.7)
The indirect similarity First perform the dilation through H by a factor of k (k ≠ 0, 1) and then the images of A, B, C are respectively X', Y', Z' with co-ordinates X'(– 2k(1 + vw), 0), Y'(– 2kvw, 2kv), Z'(– 2kvw, 2kw). Since H is the origin we may suppose the axis of reflection through H has equation y = mx. We quote the result that the image of the point with co-ordinates (c, d) after reflection in this line has co-ordinates {1/(1 + m2)}(2dm + c(1 – m2), 2cm – d(1 – m2)). Using the co-ordinates of X', Y', Z' as c, d we obtain the co-ordinates of the images X, Y, Z as X: {1/(1 + m2)}(– 2k(1 – m2)(1 + vw), – 4km(1 + vw)), Y: {2kv/(1 + m2)}(2m – w(1 – m2), – 2mw – (1 – m2)), Z: {2kw/(1 + m2)}(2m – v(1 – m2), – 2mv – (1 – m2)). The perspective We now prove that triangles ABC and XYZ are in perspective. Since this is true for all values of k ≠ 0, 1 and for all values of m this is a very general result that implies that any axis through H may be used for the reflection and any enlargement factor (k = 1would mean that AX, BY, CZ are parallel). An interesting case that we do not provide separate analysis for is when the axis through H is the Euler line and the enlargement factor is 0.5. The image of the circumcircle is then the nine-point circle and the points X, Y, Z are additional natural points on the nine-point circle. The equations of AX, BY, CZ turn out to be 2kmx + (1 + m2 – k(1 – m2))y + 4km(1 + vw) = 0,
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(8.8)
(1 + m2 + k(1 + 2wm – m2))x + (w(1 + m2) – k(w – 2m – wm2)y = 4kv(m(1 – w2) – w(1 – m2), (8.9) 2 2 2 2 2 (1 + m + k(1 + 2vm – m ))x + (v(1 + m ) – k(v – 2m – vm )y = 4kw(m(1 – v ) – v(1 – m2), (8.10) respectively. These lines are concurrent at the point Q with co-ordinates Q: (4k/{(1 – k2)(1 + m2)2})((k(m4vw + m3(v + w) + 2m2 – m(v + w) + vw) + m4vw + m3(v + w) + m(v + w) – vw), – m(k(m2(vw – 1) + 2m(v + w) – vw + 1) + (1 + m2)(1 + vw))). (8.11) Of significance is the equation of circle XYZ and this is (1 + m2)(x2 + y2) – 2k(m2(3vw + 1) + 2m(v + w) – (1 + 3vw))x – 2k(m2(v + w) – 2m(3vw + 1) – (v + w))y + 8k2vw(1 + vw)(1 + m2) = 0. (8.12) Triangle UVW and its properties We define points U, V, W as the points of intersection of the lines XH, YH, ZH with circle XYZ. This means that UVW is inscribed in the same circle as triangle XYZ and is in perspective with it. Since X, Y, Z are the images of A, B, C in the indirect similarity and H is the fixed point of the similarity, it follows that if AH, BH, CH meet the circumcircle at points D, E, F respectively, then U, V, W are the images of D, E, F in the similarity and consequently triangles DEF and UVW are similar. The equation of XH is (1 – m2)y = 2mx.
(8.13)
This meets circle XYZ again at U, with co-ordinates {4k/(1 + m2)}(– ( 1 – m2), –2m). The equation of YH is (m2w + 2m – w)y = (m2 – 2mw – 1)x. (8.14) This meets circle XYZ again at V, with co-ordinates {4kw(1 + vw)/(1 + m2)(1 + w2)}(m2w + 2m – w, m2 – 2mw – 1). The equation of ZH is (m2v + 2m – v)y = (m2 – 2mv – 1)x.
(8.15)
This meets circle XYZ again at W, with co-ordinates {4kv(1 + vw)/(1 + m2)(1 + v2)}(m2v + 2m – v, m2 – 2mv – 1). The equation of AH is y = 0. This meets the circumcircle at the point D, with co-ordinates (– 4vw, 0).
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(8.16)
The equation of BH is wy + x = 0.
(8.17)
This meets the circumcircle at the point E, with co-ordinates {4w(1 + vw)/(1 + w2)}(– w, 1). The equation of CH is vy + x = 0.
(8.18)
This meets the circumcircle at the point F, with co-ordinates {4v(1 + vw)/(1 + w2)}(– v, 1). Having defined U, V, W in the above manner their main property is that the line through D parallel to AX passes through U, the line through E parallel to BY passes through V and the line though F parallel to CZ passes through W. The line AX has equation given by (8.8), so the line parallel to AX through D has equation 2kmx + (1 + m2 – k(1 – m2))y + 8kmvw = 0.
(8.19)
It is now straightforward to show this meets the line XH, with equation (8.13), at the point U with co-ordinates {4k/(1 + m2)}(– ( 1 – m2), –2m). Similarly the line parallel to BY through E meets YH at V and the line parallel to CZ through F meets ZH at W.
References 1. H.A.W. Speckman, Perspectief Gelegen, Nieuw Archief, (2) 6 (1905) 179 – 188. 2. K. Hagge, Zeitschrift für Math. Unterricht, 38 (1907) 257-269. 3. A.M. Peiser, The Hagge circle of a triangle, Amer. Math. Monthly, 49 (1942) 524527. 4. C. J. Bradley and G. C. Smith, Hagge circles and isogonal conjugation, Math. Gaz., 91 (2007) p202. 5. C. J. Bradley and G.C. Smith, On a construction of Hagge, Forum. Geom. 7(2007) 231-247.
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Article 2 The Four Hagge circles Christopher J Bradley
A
H2 H3 G2
N3 G3
H4
G4 N4
O
N2
H
D
GN M G1
B
N1
C H1
Fig. 1
1. Preliminary analysis 1
In later sections we state and prove a number of theorems about the four Hagge circles, with respect to an appropriately chosen centre of inverse similarity P, of the four triangles BCD, ACD, ABD, ABC that comprise a cyclic quadrilateral ABCD. It is desirable therefore to give a description of the basic configuration involved and to prove some straightforward facts that are required later. We use vectors with origin O, the centre of the circumscribing circle, and with A, B, C, D having vector positions a, b, c, d each of which has magnitude R, the radius of the circle. The triangles BCD, ACD, ABD, ABC are denoted by Δ1, Δ2, Δ3, Δ4 respectively. Suffices are used for key points in these triangles. Thus Hk is the orthocentre of Δk, Nk is the nine-point centre of Δk and Gk is the centroid of Δk, k = 1, 2, 3, 4. The vector positions corresponding to the points are h1 = b + c + d, h2 = a + c + d, h3 = a + b + d, h4= a + b + c and, as in any triangle, nk = ½hk and gk = ⅓hk. The following facts are now easy to establish: (i) AH1, BH2, CH3, DH4 are concurrent at the point N with vector position ½(a + b + c + d); (ii) AN1, BN2, CN3, DN4 are concurrent at the point G with vector position ⅓(a + b + c + d); (iii) AG1, BG2, CG3, DG4 are concurrent at a point M with vector position ¼(a + b + c + d); (iv) H1, H2, H3, H4 lie on a circle of radius R with centre H, which has position vector (a + b + c + d) and moreover H1H2H3H4 is the image of ABCD under a180° rotation about N; (v) N1, N2, N3, N4 lie on a circle of radius ½R with centre N and N1N2N3N4 is homothetic with H1H2H3H4, with centre O and enlargement factor ½; (vi) The nine-point circles of triangles Δ1, Δ2, Δ3, Δ4 , each having radius ½R therefore all pass through N; (vii) A known result is that a rectangular hyperbola through the vertices of a triangle passes through it orthocentre. It follows that the rectangular hyperbola Σ through A, B, C, D passes through all of H1, H2, H3, H4; (viii) Another known result is that the centre of a rectangular hyperbola passing through the vertices of a triangle lies on its nine-point circle. In consequence of (vii) the centre of Σ is the point N; (ix) G1, G2, G3, G4 lie on a circle of radius ⅓R with centre G, which has position vector ⅓ (a + b + c + d); (x) The points O, M, G, N, H are collinear and if O and H are given co-ordinates 0, 1 on this line, then M, G, N have co-ordinates ¼, ⅓, ½ respectively. See Fig. 1 for an illustration of these results.
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2.
Setting the scene
It is clear from Speckman’s work on indirect similar perspective triangles that if we now choose any point P on the rectangular hyperbola defined in Section 1 Result (vii), it is possible to find the Hagge circle of P with respect to each of the four triangles Δ1, Δ2, Δ3, Δ4 and to draw them on the same diagram. These are the four Hagge circles Γ1, Γ2, Γ3, Γ4 of the article heading and they are shown in Fig. 2. Note that as the axes of inverse similarity through P are parallel to the asymptotes of the rectangular hyperbola Σ, these axes coincide for each of the Hagge circles. A consistent notation is essential and this we now describe. See Fig. 2. The centres of the circles are denoted by Qk, k = 1, 2, 3, 4. For reasons that become clear shortly we re-label the four orthocentres A1, B2, C3, D4, so that, for example D4 is the orthocentre of triangle ABC. The lines AP, BP, CP, DP meet the circumcircle Γ at points A', B', C', D'. We now consider the labelling of points on the Hagge circle Γ4, which is the Hagge circle of P with respect to triangle ABC. They all carry the subscript 4. The reflection of A' in BC we denote by A4', the reflection of B' in CA we denote by B4' and the reflection of C' in AB we denote by C4'. These replace the labels U, V, W used in Article 1. The points where A4'P, B4'P, C4'P, D4P meet Γ4 are denoted by A4, B4, C4, D4' respectively. The first three of these replace the labels X, Y, Z in Article 1. Points with suffices 1, 2, 3 are similarly defined on circles Γ1, Γ2, Γ3. The labels Pg1, Pg2, Pg3, Pg4 are given to the isogonal conjugates of P with respect to triangles Δ1, Δ2, Δ3, Δ4 respectively. The double lines of inverse symmetry through P are denoted by L and L'. The centre of the rectangular hyperbola Σ is denoted by M. In Section 3 we use Cartesian co-ordinates, origin M and the asymptote parallel to L is chosen as the x-axis. In this way Σ has equation xy = 1 (by choice of scale), and we use a parameter t on the hyperbola so that its points have co-ordinates (t, 1/t). The points A, B, C, D, P are given parameters a, b, c, d, p, where it is known, since ABCD is cyclic, that abcd = 1 and the parameters of A1, B2, C3, D4, since they are the orthocentres, have parameters – a, – b, – c, – d respectively.
The following theorems now hold: Theorem 1 The centres, Q1, Q2, Q3, Q4, of the four Hagge circles are collinear. Theorem 2 The points A1, A2, A3, A4, A1', A2', A3', A4' are collinear and similarly for the points Bk, Bk', k = 1, 2, 3, 4 and Ck, Ck', k = 1, 2, 3, 4 and Dk, Dk', k = 1, 2, 3, 4.
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Theorem 3 The quadrilateral Pg1Pg2Pg3Pg4 is similar to the quadrilateral A'B'C'D'.
L'
C3 B2 C2' L C4
Pg2
Pg3
D3''
A3'
D To D3' D4' D2
A
C'
B'
To A3' A3
B4'
B1 A4
C1' D1 Q3
B3'
A1'
M Q1
Q4
Q2
P
A1 B1'
D1' O
C1
D4 D3
To A3' A4'
A2
ToD2'
Pg1 C4''
B4
Pg4
B A3'
B2'
C C3' D2'
To C2
B3 A'
D'
C2
Fig. 2
3.
The analysis
The co-ordinates of O The chord AB has equation aby + x = a + b.
4
(3.1)
The midpoint of AB has co-ordinates (½(a + b), ½(a + b)/(ab)). It follows that the perpendicular bisector of AB has equation 2ab(y – abx) = (a + b)(1 – a2b2). (3.2) The perpendicular bisector of BC has equation 2bc(y – bcx) = (b + c)(1 – b2c2). These meet at O, the centre of Γ, at the point with co-ordinates (½(a + b + c + d), ½(1/a + 1/b + 1/c + 1/d). Here we have used abcd = 1. It can be shown that the radius of the circumcircle is ½√(a2 + b2 + c2 + d2 + 1/a2 + 1/b2 + 1/c2 + 1/d2).
(3.3)
The equation of the circumcircle Γ From the above information this is easily proved to be 2x2 – 2x(a + b + c + d) + 2y2 – 2y(1/a + 1/b + 1/c + 1/d) + (ab + ac + ad + bc + bd + cd + 1/ab + 1/ac + 1/ad + 1/bc + 1/bd + 1/cd) = 0.
(3.4)
Alternatively, in terms of a, b, c only it has the form abcx2 – (abc(a + b + c) + 1)x + abcy2 – (a2b2c2 + ab + bc + ca)y + (abc)(ab + bc + ca) + a + b + c = 0.
(3.5)
A digression The reflection of the point with co-ordinates (f, g) in the line with equation lx + my = n is the point with co-ordinates (h, k) where h = {f(m2– l2) + 2l(n – gm)}/(l2 + m2), k = {2mn + g(l2 – m2) – 2flm}/(l2 + m2). This is straightforward and is left to the reader. This is used later to reflect ABCD in the line L and then dilate through P to obtain the Hagge circle A4B4C4D4. The indirect similarity We now determine the equations governing the indirect similarity which maps the circumcircle into the Hagge circle of P with respect to Δ4. This is done by following what happens to the point D(d, 1/d). We know the image of this point is the orthocentre D4(– 1/abc, – abc) and we know the mapping is effected by a reflection in the line L, with equation y = 1/p followed by an 5
enlargement (reduction) by a factor PD4/PD. As far as the reflection is concerned we may use the analysis of the last paragraph with l = 0, m = 1, n = 1/p. The result of this reflection on D is to produce the point with co-ordinates (d, 2/p – 1/d). The enlargement (reduction) factor (comparing x-co-ordinates of the points concerned) is equal to (p + d)/(p – d) = (abcp + 1)/(abcp – 1). It may now be checked that the reflection in L followed by the dilation with this enlargement (reduction) factor takes the point with coordinates (h, k) to the point with co-ordinates (x, y) where x = (abchp – 2p + h)/(abcp – 1), (3.6) and y = (abc(kp – 2) + k)/(1 – abcp). (3.7) Using h = a, k = 1/a we deduce the co-ordinates of A4 to be ((a2bcp + a – 2p)/(abcp – 1), (2a2bc – abcp – 1)/(a(abcp – 1))), with similar expressions for the co-ordinates of B4 and C4 by using b and c, in x and y above, instead of a. The equations of the Hagge circles From here it is quite an elaborate calculation to obtain the equation of the Hagge circle A4B4C4 and check the fundamental theorem that D4 lies on this circle. The computer algebra package DERIVE was used to perform the calculations, and the result is that Γ4 has equation abc(abcp – 1)(x2 + y2) – (a2b2c2p(a + b + c) + abc(a + b + c) – 3abcp + 1)x + (a3b3c3p – 3a2b2c2 + (abcp + 1)(bc + ca + ab))y – 2a3b3c3 + a2b2c2p(a + b + c) + abc(bc + ca + ab) – (abcp + 1)(a + b + c) – 2p = 0. (3.8) The equations of the other Hagge circles follow immediately by using other triplets of parameters instead of a, b, c. The co-ordinates of the centres of these circles may now be written down and it has been checked, using DERIVE, that any three of the four centres are collinear. Theorem 1 is now proved, but it also follows from a geometrical argument. The centre O of the circumcircle is the common circumcentre of all four triangles Δ1, Δ2, Δ3, Δ4. If we now reflect O in the two lines of inverse similarity, it maps into a pair of points collinear with P. Since the centres of the four Hagge circles are now the images of these points by means of dilations through P with enlargements (reductions) using different factors, the resulting images all lie on this line, and this line also passes through P. Theorem 2 follows by a similar argument, bearing in mind that we already know primed and unprimed pairs of points such as A4, A4' are collinear with P. The quadrilateral A'B'C'D' The equation of the line AP is 6
x + apy = a + p.
(3.9)
This meets the circumcircle again at the point A' with co-ordinates (x, y) where x = {ap2(abc(b + c) + 1) – p(a2b2c2 + a(b + c) – bc) + abc}/{bc(a2p2 + 1)}, (3.10) y = {abcp2 + p(a2bc – abc(b + c) – 1) + ab2c2 + b + c}/{bc(a2p2 + 1)}. (3.11) The point D' has co-ordinates similar to these, but with d replacing a. From these co-ordinates we can work out (A'D')2 and the result is {(a – d)2(b – p)2(c – p)2(b2c2 + 1)}/{b2c2(a2p2 + 1)(d2p2 + 1)}.
(3.12)
The quadrilateral Pg1Pg2Pg3Pg4 The co-ordinates (h, k) of Pg4, the isogonal conjugate of P with respect to triangle ABC are given by h = (a + b + c – p)/(1 – abcp), (3.13) k = (p(bc + ca + ab) – abc)/(abcp – 1). (3.14) This may be checked as follows: Let the line from A to (h, k) meet Γ at A'', then it is easy to show that A'A'' is parallel to BC. The symmetry of h, k with respect to a, b, c now proves that (h, k) are the co-ordinates of Pg4, as this calculation reflects one of the standard constructions for an isogonal conjugate. Indeed the parallel A'A'' to BC indicates that AP and APg4 are reflections of each other in the internal bisector of angle A. The co-ordinates of Pg1 now follow from (h, k) by exchanging a and d. From these co-ordinates we can work out (Pg1Pg4)2 and the result is {(a – d)2(b2c2 + 1)(b2p2 + 1)(c2p2 + 1)}/{(abcp – 1)2(bcdp – 1)2}.
(3.15)
The similarity of the quadrangles The ratio of the squares of the side lengths and diagonals of the quadrangles Pg1Pg2Pg3Pg4 and A'B'C'D' may now be calculated and is the totally symmetric expression {(a2p2 + 1)(b2p2 + 1)(c2p2 + 1)(d2p2 + 1)}/{(a – p)2(b – p)2(c – p)2(d – p)2}. (3.16)
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Article 3 Generalizations of Hagge’s Theorems Christopher J Bradley 1. Introduction There are two unrelated generalizations of the Hagge configuration. The first generalization involves moving from the orthocentre H to another point K at the expense of losing the indirect similarity. It also appears that the material of Article 2 cannot be completely generalized, as there is a restriction on the facility for creating four circles from a cyclic quadrilateral. Also it does not appear there is a prescription that relates the point of perspective P between the triangles XYZ and UVW and the centre Q of the circle through K on which they lie. In the case of the Hagge construction Peiser [1] showed that the isogonal conjugate Pg of P is located in such a way that the nine-point centre N is the midpoint of QPg. In Sections 3 – 6 we give proofs of the main results that form the extensions of Hagge’s theorems and the four Hagge circle property. In the Hagge construction AP, BP, CP are drawn to meet the circumcircle Γ at points D, E, F respectively. These points are then reflected in the sides BC, CA, AB to create three new points U, V, W. The first Hagge theorem is that the circle UVW passes through the orthocentre H. Next one draws AH, BH, CH to meet this circle at X, Y, Z respectively and the second Hagge theorem is that UX, VY, WZ are concurrent at P. It is interesting that in this generalization one does not start with a point P, but with the centre Q of the proposed circle UVWK. The second generalization is designed to maintain the property that triangles ABC and XYZ have orthologic points with respect to each other and this leads to these triangles and their circumcircles being related by an indirect similarity and also to triangles DEF and UVW being related by the same indirect similarity. What is lost is the reflection property mentioned in the last paragraph. We describe these generalizations in turn and the following results hold for the first: 2. Results for the first generalization Theorem 1 Let ABC be a triangle and K any point not lying on its sides or extensions. With centre any point Q draw a circle Σ to pass through K. The intersections with Σ of circles BKC, CKA, AKB are denoted by U, V, W respectively. Then 1
(BU/CU)(CV/AV)(AW/BW) = – 1.
(2.1)
In this expression (BU/CU), for example, is taken as positive if KBUC is convex and negative otherwise. The converse is also true, that if U, V, W lie on a circle and this relation holds, then the circle passes through K. This result is due to Boreico [2]. Theorem 2 Given the configuration of Theorem 1, suppose now AK, BK, CK meet Σ at X, Y, Z respectively, then UX, VY, WZ are concurrent at a point P. In other words the pairs (U, X), (V, Y), (W, Z) are in involution on Σ, by means of a perspective point. These results are shown in Fig. 1.
3.
Proof of Theorem 1
Invert with respect to K and denote inverse points by primes. We require U', V', W' collinear, which is the case by Menelaus’ Theorem if, and only if, (B'U'/C'U')(C'V'/A'V')(A'W'/B'W') = – 1. However, B'U'/C'U' = (BU/CU)(KB'/KC'). Multiplying three such relationships we get the required result. 4. Proof of Theorems 2, 3 and 4 Theorem 2 We use Cartesian co-ordinates with K as origin and let Σ have centre Q with co-ordinates (– ½, – ½). This merely imposes a scale and a direction on the configuration, so there is no loss of generality. Then Σ has equation x2 + y2 + x + y = 0. (4.1) We parameterize Σ by taking lines through K to be of the form y = mx and then m serves as the parameter of the point where this line meets Σ again. Circle BKC and Σ meet on the line KU with equation (p – 1)x + (q – 1)y = 0, so the parameter for the point U is a = – (p – 1)/(q – 1). Similarly the parameters for V and W are respectively c = – (s – 1)/(t – 1) and e = – (u – 1)/(v – 1). Now circles CKA and AKB meet at A and K, so their common chord is the line AK, which therefore has equation (s – u)x + (t – v)y = 0. It follows that the parameter of X is b = – (s – u)/(t – v). Similarly Y and Z have parameters that are respectively d = – (p – u)/(q – v) and e = – (p – s)/(q – t). 2
A V
Z
X P U
K Y
W C
B
Fig. 1
To prove that UX, VY, WZ are concurrent it is sufficient to show that the pairs (a, b), (c, d), (e, f) are in involution. This is because pairs in involution on a conic must arise from a vertex of perspective. Since an involution of pairs (h, k) is specified by an equation of the form lhk + m(h + k) + n = 0 for some real numbers l, m, n (m2 ≠ nl) it follows that the pairs are in involution if
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the determinant with rows(ab, a + b, 1)(cd , c + d, 1), (ef, e + f, 1) vanishes. DERIVE verifies this is indeed the case. Theorem 3 Let VW meet AKX at L, with M, N similarly defined, then L, M, N, P are collinear. Proof Consider the hexagon VWUXKY on the circle Σ. VW^XK = L, WU^KY = M and UX^YV = P. It follows by Pascal’s theorem that L, M, P are collinear. Similarly M, N, P are collinear. The above theorems may be generalized even further, as the following theorem shows: Theorem 4 Let ABC be a triangle and D, E, F any three generally situated points. Draw any conic Σ through D, E, F. Let conic BCDEF meet Σ at U, conic CADEF meet Σ at V and conic ABDEF meet Σ at W. Further let AD, BD, CD meet Σ at X, Y, Z respectively, then UX, VY, WZ are concurrent. Proof Project E, F to the circular points at infinity. Theorem 4 is illustrated in Fig. 2
4
A
V D Y P
E
W C
B
U
F Z
X
Fig. 2
5. Restricted generalization of the four Hagge circle configuration We refer back to Article 2 for a detailed description of this configuration, but in brief what happens is as follows: Take a triangle ABC inscribed in a circle Γ. Let its orthocentre be D4. Take a point P anywhere (not on Γ or the sides of ABC), that will serve as a centre of inverse similarity. Draw the rectangular hyperbola through A, B, C, D4 and P. Its centre, as is well known, is a point M on the nine-point circle of ABC. Now perform a 180° rotation of A, B, C, D4 about M to get the points A1, B2, C3, D. Then, wherever the initial selection of P is made, it is always the case that D lies on Γ and the points A1, B2, C3 are the orthocentres of triangles BCD, ACD, ABD respectively. The four Hagge circles of P with respect to the triangles BCD, ACD, ABD, ABC can now be drawn using the same inverse spiral symmetry about P in each case. If we now choose D4 to be a general point of triangle ABC other than the orthocentre, the same construction does not work. If P is chosen anywhere (not on Γ or the sides of ABC), then it turns 5
out that the point D does not lie on Γ. In fact there is only one conic through A, B, C, D4 for which this turns out to be true. The point M that is the centre of this conic lies somewhere on the circular locus of points for which D4M produced to meet Γ at D is such that D4M = MD. P is restricted to lie anywhere on this conic, which may be a hyperbola or an ellipse. The next question to ask is how to draw the four circles that have the point P as their involution point. As we no longer have either Peiser’s [1] prescription or the inverse spiral symmetry to rely upon, there has to be some other method of obtaining the four circle centres from P. We draw on the property of the four Hagge circle configuration to provide the answer. In that configuration it turned out that points Ak (k = 1, 2, 3, 4) lie on a line through P. We call this the A-line. Similarly there is a B-line, a C-line and a D-line. We insist for the generalization that these lines must exist. We now describe how to obtain the points A4, B4, C4. A4 is the intersection of A1P with AD4, B4 is the intersection of B2P with BD4 and C4 is the intersection of C3P with CD4. The circle Σ4 is now defined to be the circle A4B4C4. In order to justify the validity of the construction certain theorems have to be proved. Theorem 5 The circle Σ4 passes through D4 and similarly Σ1, Σ2, Σ3 pass through A1, B2, C3 Theorem 6 Let circle BCD4 meet Σ4 at A4' then A4A4' passes through P. Similarly if circle CAD4 meets Σ4 at B4' then B4B4' passes through P etc. Altogether there must be twelve such lines passing through P, three for each of the circles Σk, k = 1, 2, 3, 4. However because of the symmetry of the configuration only one such case needs to be established. Theorem 7 The centres Qk of the circles Σk, k = 1, 2, 3, 4 are collinear. Their radii are in proportion to their distances from P. In Fig. 3 we illustrate all the above properties.
6
B2
C3
D2'
D line
C2' B3'
D A C1
D3 B1'
A3'
Q3 A2
M C4'
Q1
A1'
B4 D4'
P
A4
Q4 D4 B4'
A4' A1 O
C4 C1'
D1'
D1
A3 A line
Q2
A2'
B1 D3'
C
C3'
B D2 B3
Q line
B line C line
B2'
C2
Fig. 3
6. Analysis and proofs Consider the circle Γ with equation 4m2(x2 + y2) + m(m2 + 1)(bdc + acd + abd + abc – a – b – c – d)x – (m2 + 1) (bdc + acd + abd + abc + a + b + c + d)y + (m2 + 1)(ab + ac + ad + bc + bd + cd) – 2(m2 – 1) = 0. (6.1) It may be verified that it contains the points (x, y) where 7
x = {1/(2m)}(t – 1/t), y = {(1/2)}(t + 1/t),
(6.2)
for t = a, b, c, d, provided abcd = 1 These define parametrically the hyperbola with equation y2 – m2x2 = 1 and the points with parameters a, b, c, d may be taken to define the points A, B, C, D, the cyclic quadrilateral inscribed in Γ. The centre M of the hyperbola is the origin and the image of ABCD under the 180° rotation about M defines the congruent quadrilateral A1B2C3D4 and moreover the points A1, B2, C3, D4 lie on the hyperbola and have parameters – a, – b, – c, – d respectively. The condition that four points on the hyperbola are concyclic is that the product of their parameters is 1. It follows that the sets of points (ABC3D4), (AB2CD4), (AB2C3D), (A1BCD4), (A1BC3D), (A1B2CD) are concyclic and that their equations may be obtained by altering signs in the equation of Γ as appropriate. The equation of the chord of the hyperbola joining points with parameters s and t is m(1 – st)x + (1 + st)y = s + t.
(6.3)
The point P is now chosen on the hyperbola with parameter p. We can now put (s, t) successively equal to (a, – d), (– a, p) and get the equations of the lines D4A and A1P. Their intersection is by definition the point A4, which has co-ordinates (x, y), where x = {(a2 + 1)(d + p) – 2a(dp + 1)}/{2am(p – d)}, y = {(a2 – 1)(d + p) + 2a(1 – dp)}/{2a(p – d)}.
(6.4) (6.5)
The co-ordinates of B4, C4 may now be obtained by using parameters b, c rather than a. Points B1, C1, D1 follow by using parameter a instead of d and b, c, d respectively instead of a. Points A2, C2, D2 follow by using parameter b instead of d and a, c, d respectively instead of a. Points A3, B3, D3 follow by using parameter c instead of d and a, b, d respectively instead of a. The equation of the line A1P is worth recording as it is what we have termed the A-line. It has equation m(1 + ap)x + y(1 – ap) = p – a.
(6.6)
Proof of Theorem 5 Take the co-ordinates of A4, B4, C4, D4 and construct a 4 x 4 matrix consisting of rows with entries (x2 + y2, x, y, 1) for each of the four points. Then take its determinant and factorize and DERIVE provides the answer .
(6.7)
Since abcd = 1 it follows that the four points are concyclic on a circle we denote by Σ4. A similar proof establishes the existence of circles Σ1, Σ2, Σ3. 8
We now determine the equation of circle Σ4. This is done using derive by the same method as in the proof of Theorem 5, but by using current co-ordinates instead of those of D4. The result is
(6.8) The line AP meets Σ4 at the point with co-ordinates (x, y) where x = – (1/(2bcm(a2p2(m2 + 1) + 2ap(m2 – 1) + m2 + 1))) x (a2bcp(m2 + 1)(b(c – p) – cp + 1) – a(b2c(c – p)(m2 +1) – b(c2p(m2 + 1) – c(3m2(p2 – 1) – p2 + 1) – p(m2 + 1)) + p(c – p)(m2 + 1)) + b(m2 + 1)(1 – cp) + c(m2 +1) – m2p – p), (6.9) y = – (1/(2bc(a2p2(m2 + 1) + 2ap(m2 – 1) + m2 + 1))) x (a2bcp(m2 + 1)(b(c – p) – cp – 1) + a(b2c(c – p)(m2 + 1) – b(c2p(m2 + 1) + c(m2(p2 + 1) – 3p2 – 3) + p(m2 + 1)) – p(c – p)(m2 + 1)) – b(m2 + 1)(cp + 1) – c(m2 + 1) + m2p + p). (6.10) When these co-ordinates are substituted into the equation of circle BD4A1C they are found to satisfy it, and hence they are the co-ordinates of the point A4' lying on the A-line and the two circles. Similar analysis provides all twelve pairs of points on circles Σ1, Σ2, Σ3, Σ4 that are in involution through the involution point P. This completes the proof of Theorem 6. The equations of the circles Σk, k = 1, 2, 3, 4 having been obtained it is now possible to find the co-ordinates of their centres. The co-ordinates of Q1, Q4, P may then be used to show these points are collinear. This is done by forming the determinant whose three rows are (x, y, 1) where (x, y) are successively the co-ordinates of the three points. The value of this determinant turns out to be (1/(32ab2c2d(abcp – 1)(bcdp – 1))) x ((a – d)(m2 + 1)(abcd – 1)(bcdp + 1)(b3c2p(m2 +1) + b2c2(cp(m2 + 1) – 2(m2(p2 – 1) + p2 + 1)) – b(2c(m2(p2 – 1) – p2 – 1) + p(m2 + 1)) – cp(m2 + 1))). (6.11) 9
This vanishes, on account of the factor (abcd – 1) and so the three points are indeed collinear. Similarly any other two circle centres are collinear with P, so we have identified the existence of a Q-line containing all the circle centres and the point P. This establishes the first part of Theorem 7. The second part of Theorem 7 is not difficult to check. The co-ordinates of all points are now known and it soon follows that triangles such as A4B4C4 and A1B1C1 are similar and that the enlargement factor is the same as PQ1/PQ4 (distances along the Q-line are signed, depending on which side of P the centres are, and this corresponds to whether the triangles are directly homothetic through P or whether a 180° twist is involved as well). See Fig. 3 again. 7. Results for the second generalization Choose any point T, not on the sides of triangle ABC or their extensions, which will act as a pseudo-orthocentre. In the original Hagge configuration H acts as a centre of perspective and as an orthologic centre of triangles ABC and XYZ. The generalization that is appropriate is to ensure that T is an orthologic centre. An arbitrary circle is now drawn through T, which will serve as the generalized Hagge circle. CABRI indicates that there are generally two positions of T on the circle for which it will also act as a centre of perspective, but the configuration shown in Fig. 4 is not one of these cases, as that is not necessary for the generalization. To force T to be an orthologic centre drop the perpendiculars from T onto BC, CA, AB and suppose these lines meet the circle through T at points X, Y, Z respectively. This forces XYZ to have an orthologic centre with respect to ABC. Also the perpendiculars mean that, as may be proved by simple angle chasing, that triangle XYZ is similar to ABC and ordering of labels show that it is indirectly similar to ABC. It follows that the orthologic centre of ABC with respect to XYZ exists and is the point J in the figure. J and T will be corresponding points in the indirect similarity that necessarily arises as a result of the orthologic property, so J will lie on the circumcircle of ABC. Standard methods may be used to find the centre P of inverse similarity and the axes of reflection (lines of inverse similarity). Only one is shown in Fig. 3.4. Having located P draw AP, BP, CP to meet the circumcircle at D, E, F respectively. Find the images U, V, W under the indirect similarity. These are bound to lie on the circle through T. Also triangles DEF and UVW are bound to be similar, as they are related by the indirect similarity.
10
J A Y F
T Q O
T'
X
B
C
W
D Z
U E
V P
J'
Fig. 4
The following theorem now holds: Theorem 8 XPU, YPV, ZPW are straight lines. In other words triangles XYZ, UVW are in perspective with centre P, or since they all lie on a circle X, U and Y, V and Z, W are pairs in an involution on the circle through T by projection through P. 11
Despite there not being a perspective, some of Speckman’s results still hold or generalize. In particular the conics ABCPJ and XYZPT will be images of each other in the indirect similarity. These conics are not necessarily hyperbolae; and even then CABRI confirms that their asymptotes are parallel only when a perspective exists between triangles ABC and XYZ. However, triangles ABC and XYZ are paralogic, with paralogic centres at the opposite ends of the diameter to T and J respectively. These are labeled T' and J' in the figure. Theorem 9 Let VW meet XT at L, with M, N similarly defined, then LMN is a straight line. Theorem 10 The midpoint conic (midpoints of AX, BY, CZ, DU, EV, FW) exists and is illustrated in Fig. 5 along with the axis LMN. Theorem 10 is, of course, a general theorem concerning six points on a pair of conics connected by an indirect similarity and needs no separate proof. Theorem 9 is unproved, but CABRI indicated. See Fig. 5.
12
J A Y F
T M
Q O
T'
X L B
C
W Z N
U E
D
V P
J'
Fig. 5
Proof of Theorem 8 This is trivial since it is a direct consequence of the indirect similarity. The points A, P and D are mapped by the indirect similarity on to points X, P and U. Since APD is a straight line, it follows that XPU is a straight line, P being the only invariant point and lines being mapped into lines. Similarly YPV and ZPW are straight lines.
13
References 1. A.M. Peiser, The Hagge circle of a triangle, Amer. Math. Monthly, 49 (1942) 524-527. 2. I. Boreico, Result attributed to him (Private communication). Flat 4 Terrill Court 12-14 Apsley Road, BRISTOL BS8 2SP.
14
ARTICLE 4 STUDIES IN SIMILARITY Christopher J Bradley 1.
Introduction
Given a scalene triangle ABC there are precisely six points J such that the angles BJC, CJA, AJB have values = 180o – A, = 180o – B, = 180o – C in some order. Three of these points are well known to be the orthocentre H and the two Brocard points and, which in this article, for reasons that will almost immediately become clear, we denote by H+ and H-. The other three points are less familiar. We denote them by aH, bH, cH and they lie on the circumference of the orthocentroidal circle (diameter GH), where the medians meet this circle (other than at the centroid G). The angles BJC, CJA, AJB at these points are shown in the Table 1 below.
J H H+ HaH bH cH
BJC
CJA
AJB
The notation used for the six points is prompted by Table 1, which reflects the group of symmetries of an equilateral triangle. Each Brocard point may be constructed by drawing three circles and these circles together with the circles BHC, CHA, AHB produce nine circles. We first show how these nine circles may be used to locate the points aH, bH, cH. We determine the equations of the nine circles and the coordinates of aH, bH, cH showing that they have the locations stated above. We also find the coordinates of the centres of these nine circles. The nine centres have the following remarkable properties. They form six triangles with each centre appearing as a vertex of two of the triangles. Each of these triangles is similar to triangle ABC, three of them being directly similar and three being indirectly similar (with vertices in the opposite order). Also the three that are indirectly similar are mutually in perspective with O, the circumcentre of ABC, as perspector. Given a triangle ABC with orthocentre H then, if we draw a general circle through H, its points of intersection X, Y, Z with the altitudes AH, BH, CH respectively, determine a triangle XYZ that is indirectly similar to triangle ABC. The circles through H are known as Hagge circles [1]. Since 1
the similarity is indirect there is a centre of inverse symmetry P and lines through P, which are the axes of inverse symmetry. If AP, BP, CP meet the circumcircle again at D, E, F and XP, YP, ZP meet the Hagge circle again at U, V, W then triangles DEF and UVW are also similar. There is an obvious connection between the vertices of the two triangles, which is that U is the reflection of D in BC, V the reflection of E in CA and W the reflection of F in AB. The reason a Hagge circle carries a triangle that is indirectly similar to ABC by means of the construction mentioned above is that the angles subtended at H by the sides BC, CA, AB are , , respectively. The proof of this is a straightforward argument using angle chasing. One might therefore hope, that if we use either of the Brocard points H+ or H- and draw a general circle through that Brocard point and find its intersections with the line segments from that Brocard point to the vertices, then these intersections might determine a triangle ZXY or YZX, which is indirectly similar to triangle ABC. This indeed proves to be the case and again there is a centre and axis of inverse symmetry, see Bradley [2]. If one follows the same procedure with any of the three points aH, bH, cH we get triangles that turn out to be directly similar to triangle ABC. Direct similarities have rather different properties from indirect similarities and these properties are the subject of a comprehensive survey by Wood [3]. We point out how our new circles and triangles fit into Wood’s scheme. 2. The nine circles and the six triangles We first consider three circles that determine H. If you reflect O in the side BC to obtain the point aA and then draw a circle centre aA of radius R (the circumradius of ABC), then this circle is the circle BHC. It is also the case that, if bB and cC are the reflections of O in CA and AB respectively, then circles centres bB and cC of radius R are respectively the circles CHA and AHB. The common point of these three circles is therefore the orthocentre H. In terms of areal co-ordinates, since the co-ordinates of H are given by (1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)), the equations of these circles are BHC: CHA AHB
a2yz + b2zx + c2xy – (b2 + c2 – a2)x(x + y + z) = 0, a2yz + b2zx + c2xy – (c2 + a2 – b2)y(x + y + z) = 0, a2yz + b2zx + c2xy – (a2 + b2 – c2)z(x + y + z) = 0.
(2.1) (2.2) (2.3)
Using the fact that the centre of a circle is the pole of the line at infinity we find the unnormalised co-ordinates of the centres are aA(– a2(b2 + c2 – a2), – a4 – c4 + a2b2 + 2c2a2 + b2c2, – a4 – b4 + 2a2b2 + c2a2 + b2c2), (2.4) 4 4 2 2 2 2 2 2 2 2 2 2 4 4 2 2 2 2 2 2 bB(– b – c + 2b c + a b + c a , – b (c + a – b ), – b – a + b c + 2a b + c a ), (2.5) cC(– c4 – b4 + c2a2 + 2b2c2 + a2b2, – c4 – a4 + 2c2a2 + b2c2 + a2b2, – c2(a2 + b2 – c2)). (2.6) Triangle aAbBcC is congruent to triangle ABC being the image of ABC under an 180o rotation about the nine-point centre N. We next consider the three circles defining the Brocard point H+, which has co-ordinates (1/b2, 1/c2, 1/a2). These are well known to be (i) the circle through A and B touching BC at B, (ii) the 2
circle through B and C and touching CA at C and (iii) the circle through C and A touching AB at A. We denote their centres by aB, bC, cA respectively. The equations of these circles are: AH+B: a2z2 + (a2 – b2)zx – c2xy = 0, (2.7) 2 2 2 2 2 BH+C: b x + (b – c )xy – a yz = 0, (2.8) CH+A: c2y2 + (c2 – a2)yz – b2zx = 0. (2.9) The unnormalised co-ordinates of their centres are aB(2c2a2, – a4 – b4 + 2a2b2 + c2a2 + b2c2, – c2(c2 + a2 – b2)), bC(– a2(a2 + b2 – c2), 2a2b2, – b4 – c4 + 2b2c2 + a2b2 + c2a2), cA(– c4 – a4 + 2c2a2 + b2c2 + a2b2, – b2(b2 + c2 – a2), 2b2c2).
(2.10) (2.11) (2.12)
We next consider the three circles defining the Brocard point H-, which has co-ordinates (1/c2, 1/a2, 1/b2). These are well known to be (i) the circle through A and C touching BC at C, (ii) the circle through B and A and touching CA at A and (iii) the circle through C and B touching AB at B. We denote their centres by aC, bA, cB respectively. The equations of these circles are: CH-A: AH-B: BH-C:
a2y2 – b2zx + (a2 – c2)xy = 0, b2z2 – c2xy + (b2 – a2)yz = 0, c2x2 – a2yz + (c2 – b2)zx = 0.
The unnormalised co-ordinates of their centres are aC(2a2b2, – b2(a2 + b2 – c2) , – c4 – a4 + a2b2 + 2c2a2 + b2c2), bA(– a4 – b4 + b2c2 + 2a2b2 + c2a2, 2b2c2, – c2(b2 + c2 – a2)), cB(– a2(c2 + a2 – b2), – b4 – c4 + c2a2 + 2b2c2 + a2b2, 2c2a2).
(2.13) (2.14) (2.15)
(2.16) (2.17) (2.18)
The co-ordinates of the nine centres are unnormalised but in each case they sum to the same amount (a + b + c)(b + c – a)(c + a – b)(a + b – c). The nine circles are illustrated in Fig. 1. Also shown in Fig. 1 is the orthocentroidal circle on GH as diameter, with equation shown in Bradley and Smith [4] to be (b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2 – a2yz – b2zx – c2 xy = 0. (2.19) The median AG meets this circle again at the point aH with co-ordinates aH(a2, b2 + c2 – a2, b2 + c2 – a2). To prove that this is indeed the point aH featured in Section 1, one may verify that it lies on each of the circles with centres aA, aB, aC and consequently the sides BC, CA, AB subtend at aH the angles , , these being the angles in the segments of those circles. Similarly bH and cH lie on the orthocentroidal circle and on the medians BG and CG respectively and have co-ordinates bH(c2 + a2 – b2, b2, c2 + a2 – b2) and cH(a2 + b2 – c2, a2 + b2 – c2, c2). The point bH lies on the circles with centres bA, bB, bC and cH lies on the circles with centres cA, cB, cC. 3. Lengths of segments and the similarities If ABC is an acute-angled triangle the distances from the vertices to the orthocentre and to the Brocard points are known, see Shail [5], to be AH = 2R cos A, BH = 2R cos B, CH = 2R cos C; AH+ = 2Ro sin B/sin A, BH+ = 2Ro sin C/sin B, CH+ = 2Ro sin A/ sin C; 3
AH- = 2Ro sin C/sin A, BH- = 2Ro sin A/sin B, CH- = 2Rosin B/sin C, where Ro = ½ abc/(b2c2 + c2a2 + a2b2)1/2. The reader may verify that the corresponding formulae for aH, bH, cH are A aH = 4Ro cos A, B aH = 2Ro sin A/sin B, C aH = 2Ro sin A/ sin C, A bH = 2Ro sin B/ sin A, B bH = 4Ro cos B, C bH = 2Ro sin B/ sin C, A cH = 2Rosin C/sin A, B cH = 2Ro sin C/sin B, C cH = 4Ro cos C, where Ro is, in each case, the radius of the pedal triangle of the corresponding point. In an obtuse-angled triangle these become signed lengths. For the point aH the value of Ro = ½bc/(2b2 + 2c2 – a2)1/2 with a similar formula for the points bH, cH.
aB
A
bA cC
O
aC
bH
H+ G aH
cA
bB
H
HcH C B bC
aA
cB
Fig. 1(a)
We next consider the triangle cA aB bC and show that it is similar to triangle ABC. 4
The displacement bC aB = (a2(a2 + b2 + c2), – a4 – b4 + c2a2 + b2c2, b4 – b2 c2 – a2b2 + 2c2a2 Using the areal metric , see Bradley [6], we find that (bC aB)2 = ka2, where k = (3a2b2c2(a2 + b2 + c2) + 2(b4c4 + c4a4 + a4b4) – a6(b2 + c2) – b6(c2 + a2) – c6(a2 + b2)) (3.1) all divided by {(a + b + c)(b + c – a)(c + a – b)(a + b – c)}2. Similarly (cA bC)2 = kb2 and (aB cA)2 = kc2, from which the result follows. Another short calculation shows that triangle bA cB aC is congruent to triangle cA aB bC. These two triangles together with ABC and aA bB cC are all directly similar to one another and fall into two congruent pairs. We now consider the three triangles aA aB aC, bA bB bC and cA cB cC. Note first that the centres aA, bC, cB all lie on the perpendicular bisector of the side BC. Similarly the centres bB, cA, aC all lie on the perpendicular bisector of CA and the centres cC, aB, bA all lie on the perpendicular bisector of AB. It follows that triangles aA aB aC, bC bA bB, cB cC cA are mutually in perspective with the common perspector O.
5
aB A bA
cC
O
aC
bH
H+ G aH
cA
bB
H
HcH C B bC
aA
cB
Fig. 1(b)
The displacement aB aC = (2a2(b2 – c2), a2(a2 – 3b2 – c2), a2(b2 + 3c2 – a2)). Using the areal metric we find (aB aC)2 = la2, where l = {a2(2b2 + 2c2 – a2)}/{(a + b + c)(b + c – a)(c + a – b)(a + b – c)}. (3.2) 2 2 2 2 Similar calculations show that (aC aA) = lb and (aA aB) = lc . It follows that triangle aA aB aC is similar to triangle ABC. In similar manner we find (bB bC)2 = ma2, (bC bA)2 = mb2 and (bA bB)2 = mc2, where m = {b2(2c2 + 2a2 – b2)}/{(a + b + c)(b + c – a)(c + a – b)(a + b – c)}, (3.3) showing that triangle bA bB bC is also similar to triangle ABC. And again in similar manner we find (cB cC)2 = na2, (bC bA)2 = nb2 and (bA bB)2 = nc2, where n = {c2(2a2 + 2b2 – c2)}/{(a + b + c)(b + c – a)(c + a – b)(a + b – c)}, (3.4) 6
showing that triangle bA bB bC is also similar to triangle ABC. All three triangles are directly similar to one another and inversely similar to triangle ABC. The areas of the three triangles are in the ratio l : m : n. The six triangles are shown in Fig.1(b).
References 1. 2. 3. 4.
K. Hagge, Zeitschrift für Math. Unterricht, 38 (1907) 257-269. C.J. Bradley, Omega Circles (Article 5 in this series). F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73. C.J.Bradley & G.C.Smith, The locations of Triangle Centres, Forum Geom., 6(2006) 5770. 5. R.Shail, Some Properties of Brocard Points, Math. Gaz., 80 (1996) 485-491.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
7
ARTICLE 5 Omega Circles Christopher J Bradley
A
F S
D O
U
X T
Y
Q
W
P R
V C
B
Z E
Figure
1. Introduction Every circle passing through H, the orthocentre of a triangle ABC is called a Hagge circle, named after the first person [1] to detail their most important properties. These are (i) If AH, BH, CH meet a Hagge circle Σ at points X, Y, Z respectively, then triangle XYZ is indirectly similar 1
to triangle ABC, (ii) If circles BHC, CHA, AHB meet Σ at U, V, W respectively, then UX, VY, WZ meet at a point P and (iii) P is also the centre of inverse similarity, meaning there is an axis through P such that if ABC is reflected in this axis followed by an appropriate enlargement (reduction) centre P, it is mapped on to triangle XYZ. It may be added that if triangle UVW is mapped back (by the inverse transformation) from Σ to the circumcircle to form a triangle DEF then DEF is inversely similar to UVW and because of property (ii) APD, BPE, CPF are straight lines. If you take a general point J rather than H and repeat the construction then the triangle XYZ is no longer similar to ABC. That might have been the end of the story, but it now appears that circles through the Brocard points Ω and Ω' have properties nearly the same as Hagge circles, but with subtle differences. To be particular we restrict our discussion for the time being to Omega circles, those passing through the Brocard point Ω, which if AB > AC is further from BC than the Brocard point Ω' and which has areal co-ordinates (1/b2, 1/c2, 1/a2). The properties of an Omega circle Γ are (i) If AΩ, BΩ, CΩ meet Γ again at Z, X, Y respectively, then triangle XYZ is indirectly similar to triangle ABC, (ii) If circles BΩC, CΩA, AΩB meet Γ at U, V, W then UZ, VX, WY meet at a point P. However, the centre of inverse similarity R is in general distinct from P, so that when the points D, E, F are obtained by finding the inverse images of U, V, W respectively, the lines AE, BF, CD meet at a point T which is the inverse image of P under the indirect similarity centre R. The exception is the seven-point circle, when R and P coincide. These properties are all exhibited in the figure above, which is drawn using CABRI II plus. There is good reason to believe that H, Ω, Ω' are the only three points that exhibit the indirect similarities, but there are three other points which provide direct similarities rather than indirect similarities between ABC and XYZ. There are numerous other properties that emerge from both Hagge circles and Omega circles such as the fact that pairs of triangles are orthologic and that the perspectives also create axes of importance. In this article, we do not attempt to catalogue all these properties. The interested reader is referred to the work of Speckman [2] and to previous articles in this series. 2. The equation of an Omega circle We use areal co-ordinates with A, B, C the triangle of reference. The point Ω has co-ordinates (1/b2, 1/c2, 1/a2). From the equations of AΩ, BΩ, CΩ we may write down possible co-ordinates of Z, X, Y which are Z(1, l/c2, l/a2), X(m/b2, 1, m/a2), Y(n/b2, n/c2, 1). However, l, m, n are related to one another because the circle XYZ passes through Ω. The condition for this is l = – a2b2c2(a2(b2c2 + n(c2 – m)) + c2m(b2 + n)) all divided by a4b2(b2c2+ n(c2 – m)) + a2c2b4(m – n) – b2m(c2 + n) – c2mn – b4c2mn. (2.1) What Equation (2.1) implies is that there are a doubly infinite number of Omega circles given by varying the values of m and n. By eliminating l in favour of m and n means that symmetry is lost. 2
Despite that further working becomes easier to check using a programme such as DERIVE, which is the one we use. The equation of the Omega circle, in terms of m and n, is b4x2(a2(b2c2 + n(c2 – m)) + c2m(b2 + n)) + b2x(y(a2(b4c2 – b2(c4 + mn) – c4n) + c2m(b4 – b2c2 – c2n)) + z(a4n(b2 + m) – a2(b4n + b2m(c2 + n) + c2mn) – b4mn)) + a2(c2my2(b2(c2 + n) + c2n) – c2yz(a2(b4 + b2(m + n) + mn) + m(b4 – b2c2 – c2n)) + b2nz2(a2(b2 + m) + b2m)) = 0. (2.2) 3. The indirect similarity The fact that triangles ABC and XYZ are indirectly similar follows by angle chasing as a result of the facts that angle BΩC = 1800 – C, angle CΩA = 1800 – A, and angle AΩB = 180o – B. (If Ω is replaced by H then C, A, B are replaced by A, B, C in these equations, involving a cyclic rearrangement of A, B, C.) 4. The circles BΩC, CΩA, AΩB The circle BΩC has equation b2x2 – xy(c2 – b2) + a2yz = 0.
(4.1)
The equations of circles CΩA, AΩB may be written down by cyclic change of x, y, z and a, b, c. 5. The points U, V, W and P If a triangle ABC is given and a point J and any circle Σ through J, it is a general result that if AJ, BJ, CJ meet Σ again in points Z, X, Y and circles BJC, CJA, AJB meet Σ again in points U, V, W then XV, YW, ZU are concurrent at a point P, so we do not repeat the proof, but simply give the co-ordinates of the points in the present context. (The result can then be checked in this case.) Note that the perspective is quite independent of the similarity in this case between ABC and XYZ. U has co-ordinates (x, y, z) where x = a2m(b2c2 + n(b2 + c2)), y = b2n(a2b2 + m(a2 + b2)), (5.1) 2 2 2 2 2 2 2 2 2 2 2 2 z = (b m(a (b n + c (m – n) + 2mn) +mn(b – c ))(b (c + n) + c n))/(n(a (b + m) + b m)). V has co-ordinates (x, y, z) where x = – (n(a2(b2 + m) + b2m)(a4(b2c2 + n(c2 – m)) – a2c2(b2(c2 – m + n) + n(c2 – m)) – c2m(b2(c2 + n) + c2n)))/(b2(a2(b2c2 + n(c2 – m)) + c2m(b2 + n))), y = n(a2(b2 + m) + b2m), (5.2) 3
z = a2(b2c2 + n(c2 – m)) + c2m(b2 + n). W has co-ordinates (x, y, z) where x = a2m(b2(c2 + n) + c2n), y = (b2(a2c2(b4 + b2(m + n) +mn) – b4mn)(a2(b2c2 + n(c2 – m)) +c2m(b2 + n)))/(c2m(b2(c2+ n) + c2n)), (5.3) 2 2 2 2 2 2 2 z = b (a (b c + n(c – m)) + c m(b + n)). P has co-ordinates (x, y, z) where x = (c2mn(a2(b2 + m) + b2m))(b2(c2 + n) + c2n), y = (b2n(a2(b2 + m) + b2m))(a2(b2c2 + n(c2 – m)) + c2m(b2 + n)), z = (b2(a2(b2c2 + n(c2 – m)) + c2m(b2 + n)))(m(b2(c2 + n) + c2n)).
(5.4)
It may be proved that VW^AΩ, WU^BΩ, UV^CΩ, P are collinear, but that we leave to the reader. As with all indirect similarities the midpoints of AX, BY, CZ, DU, EV, FW lie on a conic. 6. The points R, T, S The points R, T, S are the additional points in the figure. In view of the indirect similarity between ABC and XYZ there are bound to be a point R and two axes through R that are the centre and axes of inverse similarity. Nothing additional is proved by working out the co-ordinates of R, as it is known to exist. Its position is constructed as follows. The directions of the axes are parallel to the bisectors of any pair of corresponding sides such as BC and YZ. Take any such line and reflect triangle ABC in it. It will now be found that AX, BY, CZ are concurrent at a point that does not lie on the line. Move the line parallel to itself and when the point lies on the line that will be R, the centre of inverse similarity. This construction, of course, cannot be done with complete precision. But the construction can be made precise by noting that R must lie on certain lines. For example the point Ω lies on the circle XYZ. Its angular position relative to these points can be mirrored precisely on the circumcircle of ABC. Call that point Ωo. Call the same Brocard point of triangle XYZ the point Ω1. The passage from Ωo to Ω1 is now the product of two indirect similarities and is thus an enlargement, so that ΩoΩ1 must pass through R. A second such point, leading to a second line through R, might be the second point of intersection of the circle XYZ with the seven-point circle. The two lines fix the position of R exactly.
4
The point T is the inverse image of the in direct similarity of the point T and if DEF is the inverse image of triangle UVW then AE, BF, CD are concurrent at T. Since circles BUC, CVA, AWB are concurrent at Ω, it follows that circle AVW, BWU, CUV are concurrent at a point S. As indicated by CABRI II plus, S lies on the circumcircle of ABC. This curious fact we have been unable to prove, as DERIVE found the equations too technically complicated to solve. 7. Omega prime circles
A
V S
Y
E X
D
Q
R
W
P O U
T Z
B
'
C
F
Figure 2
5
Circles through Ω' (1/c2, 1/a2, 1/b2) have very similar properties to those passing through Ω, the difference arising from the labelling of the points X, Y, Z. Now AΩ', BΩ', CΩ' meet the Omega prime circle at points Y, Z, X respectively. Then there is an indirect similarity between triangles ABC and XYZ. This similarity follows again by simple angle chasing bearing in mind that angle BΩ'C = 180o – B, angle CΩ'A = 180o – C and angle AΩ'B = 180o – A. 8. aH bH cH circles
A
A' J
X
V F'
bH
O
E'
G
Q B' F W
E
H S D' B
Z
C
Y D
U
C' K
T
P
F IGURE 3
aH, bH cH are the intersections of the lines AG, BG, CG with the orthocentroidal circle on GH as diameter, where G is the centroid of ABC and H is its orthocentre. We illustrate in Figure 3 what happens when a circle Σ is drawn through bH. The lines AbH, BbH, CbH meet Σ again at
6
Z, Y, X respectively. Since angle BbHC = 180o – C, angle CbHA = 180o – A, angle AbHB = 180o – B it follows by simple angle chasing that triangle XYZ is directly similar to triangle ABC. The intersections of Σ with the circumcircle are denoted by J and K. According to the theory of Wood [3] either J or K may be used as a perspector to project triangle XYZ into a triangle A'B'C', where A' is the intersection of JX with the circumcircle and B' and C' are similarly defined. Triangle A'B'C' is directly similar to XYZ and so directly congruent to ABC. It may be seen in the figure that they are related by a rotation about O. If Σ is not large enough to intersect the circumcircle it may be enlarged about its centre Q until it does. All that happens is that an extra step is required to illustrate the direct similarity. Once this has been done the rest follows as with Omega circles. Points U, V, W are defined as intersections with Σ of the circles BbHC, CbHA, AbHC respectively and UZ, VY, WX intersect at P. Points D, E, F are the inverse images of U, V, W and DC, EB, FA intersect at T. The point S is the intersection of circles AYZ, BZX, CXY and it also lies on the circumcircle. References 1. K. Hagge, Zeitschrift für Math. Unterricht, 38 (1907) 257-269. 2. H.A.W. Speckman, Perspectief Gelegen, Nieuw Archief, (2) 6 (1905) 179 – 188. 3. F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
7
Article 6 On The Ten Point Wood-Desargues’ Configuration Christopher J Bradley and Geoff C Smith Abstract The ten pairs of directly similar triangles in perspective in the Wood-Desargues’ configuration have ten common Hagge circle centres. These centres are known to lie one each of the ten perspectrices. It is shown in this paper that these centres also lie at the vertices of five cyclic quadrangles, which are similar to the five cyclic quadrangles of the initial Wood-Desargues’ configuration and that these five new cyclic quadrangles have a common circumradius. 1.
Introduction
The Wood-Desargues’ configuration is built up as follows: Start with two circles intersecting at points J and K. In the first circle inscribe a triangle ABC. Draw AK, BK, CK to meet the second circle at points a, b, c. It follows that triangles ABC and abc are directly similar triangles in perspective with J the centre of direct similarity. By this we mean that triangle abc may be obtained by from triangle ABC by a rotation about J followed by a dilation centre J. Suppose that BC meets bc at the point 1, CA meets ca at the point 2 and AB meets ab at the point 3. Then 123 is the Desargues’ axis of perspective, commonly referred to nowadays as the perspectrix. There are in all ten points A, B, C, K, a, b, c, 1, 2, 3 which form what may be called the Wood-Desargues’ configuration. The ten points turn out to be equivalent in the following sense. Each of them is a vertex of perspective of two triangles in the configuration, the remaining three points being collinear and serving as the perspectrix. This situation is illustrated in Figure 1. In his landmark paper Wood [1] proved that the quadrangles Aa23, Bb31, Cc12 are cyclic in addition to the starting quadrangles ABCK and abcK. These circles are shown in Figure 1 and for reasons, which will emerge shortly, their centres are also labelled ABCK, abcK etc. In his paper Wood showed that the five circle centres are themselves concyclic and lie on a circle passing through J. The ten pairs of triangles in perspective, together with their vertices of perspective and their perspectrices are shown in Table 1. The orders of the vertices in pairs of triangles correspond and the order of letters in the perspectrix is standard, so, for example 1, is the intersection of BC and bc, BC and 32 and bc and 32. Triangle 1 ABC KBC AKC ABK Cc2
Triangle 2 abc a32 3b1 21c Bb3
Vertex K A B C 1 1
Perspectrix 123 1cb c2a ba3 aAK
Aa3 Bb1 Kbc Kca Kab
Cc1 Aa2 A32 B13 C21
2 3 a b c
bBK cCK 1CB 2AC 3BA
Table 1: The ten perspectives
In Section 2 we consider the orthocentres of the twenty triangles and show that they lie four at a time on five quadrangles similar to the initial five quadrangles and also that the same twenty orthocentres lie four at a time on five lines. This follows from well-known results about cyclic quadrangles and quadrilaterals, but is presented, as the result is required in what follows. In Section 3 we establish our main results. Each pair of triangles in the list of ten perspective pairs have orthocentres, say H and F, so that the circle JHF is a common 2
Hagge circle of the two triangles. Its centre we denote by h, and it is the intersection of the perpendicular bisectors of JH and JF. It is known that these ten points h lie one each on the ten perspectrices, see [1]. We show that they also form the vertices of five cyclic quadrangles similar to the initial five cyclic quadrangles, and that each of the five circles involved has the same circumradius. The method of proof involves the pentagon formed by the five circle centres of the initial five circles. The result follows from lemmas that are proved and from straightforward extensions of the work of Wood [1]. 2. The twenty orthocentres In Figure 2 we show the circle ABCK and its four component triangles ABC, KBC, AKC, ABK with orthocentres H(K), H(A), H(B), H(C) respectively. In the same figure we also show the four lines 123, b1c, ac2, 3ba, which are the four perspectrices of the four perspectives associated with the triangles of the cyclic quadrangle ABCK. The four orthocentres of the triangles formed by these four lines, namely triangles abc, a32, 3b1 and 21c (see Table 1) are denoted by F(K), F(A), F(B), F(C) respectively. Proposition 1 The quadrangle H(A)H(B)H(K)H(C) is congruent to the quadrangle ABKC. Proof The result is true for any cyclic quadrangle and does not depend upon A, B, C, K being part of the Wood-Desargues’ configuration. Let the centre of the circle ABCK be the origin of vectors and suppose A, B, C, K have vector positions a, b, c, k respectively. Then H(K) has vector position a + b + c and H(A) has vector position b + c + k. It follows that H(A)H(K) = a – k = KA. It is evident that H(A)H(B)H(K)H(C) is the image of ABKC under a 180o rotation about the point whose vector position is ½ (a + b + c + k). Proposition 2 The points F(A), F(B), F(C), F(K) are collinear. Proof The result is true for the orthocentres of the four triangles formed by any set of mutually non-parallel lines and does not depend on the lines 123, b1c, ac2, 3ba being part of the Wood-Desargues’ configuration. The result is immediate from the work of Steiner on the complete quadrilateral. See, for example, Durrell [2]. Now there are five circles and associated quadrangles in the initial Wood-Desargues’ configuration, so there are similar results to Propositions 1 and 2 for each of the other four. This means that each of the twenty orthocentres of the twenty triangles of the ten perspectives appears twice. Each will appear once as an H point of one of the five quadrangles and once as an F point of one of the five lines. The five resulting cyclic quadrangles are congruent in each case to one of the initial five quadrangles.
3
c 1 F(C) J F(B)
A h(B)
F(K)
h(A) b
C H(B)
2 h(K)
H(K) K B
h(C)
H(C)
a H(A) 3 F(A)
Figure 2
The perpendicular bisectors of JH(K) and JF(K) meet at a point we denote by h(K). This is the joint Hagge centre of triangles ABC and abc. It is known to lie on the perspectrix 123. The points h(A) and h(B) and h(C) are similarly defined for the pairs of triangles KBC, a32 and AKC, 3b1 and ABK, 21c respectively and lie on the lines b1c, ac2, 3ba respectively. In Figure 2 we also show these points and the quadrangle h(A)h(B)h(K)h(C). There are five such quadrangles and each of the h points appears on two of them. For example h(K) lies on the quadrangle h(a)h(b)h(c)h(K) as well. In Section 3 we show that these five quadrangles are directly similar to the five initial quadrangles in the WoodDesargues’ configuration.
4
3. The ten common Hagge centres
First we establish a very simple Lemma that at first sight does not appear to be related the Wood-Desargues’ configuration at all, but is in fact a useful step in establishing two of the theorems we want to prove. The Lemma is illustrated in Figure 3. Lemma 1 Suppose that P, Q, R are three non-collinear points and lines p, q, r through P, Q, R respectively meet at a point S on circle PQR. Then lines p , q , r through P, Q, R respectively, perpendicular to p, q, r respectively, are concurrent at a point T diametrically opposite to S. Proof Let the lines p and q meet at T. Then, since SPT = SQT = 90o, T lies at the other end of the diameter to S of circle SPQ. But R lies on this circle, and r and r are at right angles. It follows that r passes through T. It is not needed here, but it is easy to verify that the condition that S lies on circle PQR is necessary as well sufficient for the concurrence of p , q , r .
5
We now establish the crucial Lemma in helping to establish what we wish to prove. One possible configuration is shown in Figure 4, but the proof given using directed angles is diagram independent.
Lemma 2 Let J, O, L be three points on the circumference of a circle S1 and let S2 and S3 be the circles through J, centres L and O respectively. Suppose that S2 and S3 meet again at A, S3 and S1 meet again at Z and S1 and S2 meet again at W, then A, L and Z are collinear as are A, O and W. Proof Figure 5 is a copy of Figure 1, but with some additional points and lines added. First, however, note that the centres of the five circles have been relabelled, so that circle ABCK has centre U, circle abcK has centre V, circle Aa23 has centre L, circle Bb13 has centre M and circle cC12 has centre N. Consider the point L, the centre of circle Aa23.
6
This is clearly the point of intersection of the perpendicular bisectors of JA, Ja, J2, J3. Now A, a; A, 2; A, 3; a, 2; a, 3, 2, 3 are six corresponding pairs of vertices amongst the thirty such pairs in the ten perspectives in Table 1. The same applies to each of the five circle centres and this fact accounts for all thirty such points of intersection. Additional points in the diagram are X, Y, Z, W. These points occur in Propositions we establish next. Lines ALZ, AUW, BMZ, CNZ are also shown, and these collinearities will soon be established. Proposition 3 Let Z be the point of intersection of circles JUVLMN and JABCK, and let W be the point of intersection of circles JUVLMN and JAa23, then ALZ, AUW, BMZ, CNZ are straight lines. Proof In any ten point Wood-Desargues’ configuration, the conditions of Lemma 2 apply, which accounts for the straight lines ALZ and AUW. Application of the same Lemma to other pairs of circles shows that BMZ and CNZ are also straight lines. 7
Proposition 4 Triangles ABC and LMN are directly similar triangles in perspective with centre of direct similarity J and vertex of perspective Z. Proof This follows immediately as a result of Proposition 3. Since circle JUVLMN cuts the five initial circles in five different points, there are thus five such vertices of perspective and all twenty triangles of the initial Wood-Desargues’ configuration get mapped by perspectives on to the triangles of the pentagon UVLMN. We have thus established Proposition 5 Consider the pentagon UVLMN. Omit one vertex, say M. Then the quadrangle UVLN is directly similar to the quadrangle Bb31 (a little care has to be taken over the order of the vertices). The same applies when any other vertex of the pentagon is omitted. In this way the pentagon inscribed in the circle UVLMN carries all the angular information of the five quadrangles of the initial configuration. If we now use Lemma 1 with X the point diametrically opposite Z on the circle JABCK we recover Theorem 12 of Wood [1], namely Proposition 6 Tangents at A, B, C to circles Aa23, Bb13, Cc12 are concurrent at a point X on circle JABCK. Proof These tangents are perpendicular to the normals AL, BM, CN so Lemma 1 applies. There are, of course other cases in which sets of tangents are concurrent. Lemma 1 also provides proof of the following proposition. Proposition 7 Lines through L, M, N parallel to the tangents at A, B, C in Proposition 6 are concurrent at a point Y on circle JUVLMN, which is diametrically opposite Z in that circle. This allows us to identify the centre of the circle JUVLMN. It is the midpoint of YZ. Recall now that h is the intersection of the perpendicular bisectors of JH and JF and this is how all ten common Hagge centres are constructed. We have shown above that if we take the perpendicular bisectors of JA and Ja we obtain L, if we take the perpendicular bisectors of JB and Jb we get M and if we take the perpendicular bisectors of JC and Jc 8
we get N. But we have shown in Proposition 5 that triangles ABC and LMN are in perspective and similar with vertex. It follows that h(K) is the orthocentre of triangle LMN. Similar considerations hold for all the other common Hagge centres, which together account for all ten orthocentres of the triangle formed by the pentagon UVLMN and its diagonals. Proposition 8 The quadrangle h(A)h(B)h(C)h(K) is similar to the quadrangle ABCK. Proof We know from Proposition 1 that the quadrangles ABCK and H(A)H(B)H(C)H(K) are congruent. For the same reason quadrangles LMNV and h(A)h(B)h(C)h(K) are congruent. But from Proposition 5 the quadrangles ABCK and LMNV are similar. There are five such cyclic quadrangles formed by the common Hagge centres and since the orthocentres of the triangles formed by UVLMN and its diagonals all derive from the same circle we have proved Proposition 9 The ten common Hagge centres of the Wood-Desargues’ configuration are inscribed in five circles of equal radius and each Hagge centre lies on two such cyclic quadrangles. The configuration is illustrated in Figure 6, in which the initial configuration and its five circles and quadrangles are shown, together with the perspectrices and the common Hagge centres lying on the perspectrices and forming five cyclic quadrangles of equal radius.
9
References 1. F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73. 2. C.V. Durell, Modern Geometry, Macmillan (1946).
10
ARTICLE 7 Generalization of the Wallace-Simson line Christopher J Bradley 1.
Introduction
If ABC is a triangle and J is a point on its circumcircle Σ, then the Wallace-Simson line property is that the feet of the perpendiculars L, M, N from J on to the sides of ABC are collinear and LMN is called the Wallace-Simson line. Furthermore, if H is the orthocentre of ABC then the midpoint Q of JH lies on LMN. If instead of taking the feet of the perpendiculars on to the sides, we take the reflections L', M', N' of J in the sides, then L'M'N' is also a straight line which we refer to as the Double Wallace-Simson line of J. As this line is a degenerate Hagge circle it is known that L'M'N' passes through the orthocentre H. In this chapter we provide a generalization of this construction, so that the Wallace-Simson line is just a particular case, being one of an infinite number of lines that we refer to as generalized Wallace-Simson lines. Later we identify these lines as ones that are already known, providing an explanation of why these lines are the same as those we describe. Figure 1 illustrates the general construction. The point Q is now varied from its fixed position at the midpoint of JH from which the well known Wallace-Simson line results and becomes a variable point on the perpendicular bisector of JH. The circle S is drawn, centre Q, passing through J and H. The points where the altitudes of ABC meet S are denoted by X, Y, Z, where AX, BY, CZ are perpendicular to BC, CA, AB respectively. Circles AJX, BJY, CJZ are then drawn and L is the point of intersection of BJY and CJZ with M, N similarly defined. As we prove in due course, the following results hold: 1. L, M, N lie on BC, CA, AB respectively; 2. L, M, N are collinear. 3. The line LMN contains the point Q. The line LMN is, of course, what we mean by a generalized Wallace-Simson line and as Q varies we get an infinite number of such lines all defined with respect to the one fixed point J on the circumcircle of ABC. Various subsidiary results emerge in the course of establishing these propositions, and we now briefly indicate the plan we adopt to prove them.
1
M Z
X K Ao A
J Q Co N Bo
H Y B
C
L
Fig. 1
The method we use is to take J to be the centre of the direct similarity that maps H to Q. We then apply this direct similarity to ABC to get a triangle AoBoCo and it transpires that Σo, the 2
circumcircle of AoBoCo, shares the point J with Σ and a second point K such that AAo, BBo, CCo concur at K. This is typical of directly similar triangles in perspective, see Wood [1]. When Q is the midpoint of JH triangle AoBoCo has its sides parallel to ABC and is half the size. In all cases it turns out that the line LMN we want is the Double Wallace- Simson line of J with respect to triangle AoBoCo and, being a Double Wallace-Simson, line it automatically passes through the orthocentre of AoBoCo, which by construction is Q. Synthetic considerations may be used to show that L, M, N lie on BC, CA, AB respectively and indeed that LMN is a straight line. The bonus of the analysis is that it highlights the role played by the direct similarity. 2.
Defining the direct similarity
We use Cartesian co-ordinates and take J to be the origin (0, 0) and the centre O of triangle ABC to the point (1, 0). Throughout the calculation the computer algebra package DERIVE was used. If, for the moment, we suppose H has co-ordinates (h, k), then it is straightforward to show that a variable point Q on the perpendicular bisector of JH has co-ordinates (½h – kt, ½k + ht), where t is a variable parameter. The direct similarity through J that maps H to Q is therefore given by the matrix (2.1)
If considered as a dilation of magnitude d with centre of enlargement J and a rotation by an angle θ about J we see that d and θ are related by the equation d = 1/(2 cosθ) and also that t = ½ tan θ. (In all that follows we suppose without loss of generality that cosθ is positive and that – ½ π < θ < ½ π.) The equation of the circumcircle Σ is x2 + y2 – 2x = 0. We suppose that the point A has co-ordinates (
(2.2) , where a is a parameter, and that
B, C have similar co-ordinates with b, c respectively replacing a. Applying the direct similarity to A we obtain a point Ao with co-ordinates
. The points Bo, Co, the
images of B, C respectively under the direct similarity have similar co-ordinates with b, c respectively replacing a. We now prove that triangle AoBoCo is in perspective with triangle ABC. The equation of AAo is (a – 2t)x – (1 + 2at)y + 4t = 0. (2.3)
3
This meets the circumcircle Σ with Equation (2.2) at A and at the point K with co-ordinates (
). Since the co-ordinates of K are independent of a, b, c it follows that AAo, BBo,
CCo are concurrent at K and the triangles are not only similar, but in perspective. From the theory of Wood [1] on directly similar triangles in perspective it follows that the circle Σo, the circumcircle of triangle AoBoCo intersects Σ at the points J and K. See Fig. 1. We now consider the circle, centre Ao passing through J. Its equation is soon found to be (1 + a2)x2 + (1 + a2)y2 – 2(1 – 2at)x – 2(a + 2t)y = 0. (2.4) It may now be checked that this circle passes through A. So this circle is now identified as circle JAX, though we have yet to find the co-ordinates of X and to show it lies on the Hagge circle, centre Q passing through H and J. If the parameter a in Equation (2.4) is replaced by b, c we have the equations of the circles that may be identified as circles JBY, JCZ respectively. Note that in view of the direct similarity triangles JAoA, JBoB, JCoC are similar. The co-ordinates of H may be obtained from those of O, A, B, C and are
(2.5) From this we may obtain the equation of the altitude AH, which is (1 + a2)(b + c)x – (1 + a2)(1 – bc)y + 2(a + b + c – abc) = 0.
(2.6)
This meets the circle centre Ao through J at the point A and the point X with co-ordinates ).
(4.2.7)
The co-ordinates of Y and Z may be found from those of X by cyclic change of the symbols a, b, c. The equation of the circle through X, Y, Z may now be calculated and is (1 + a2)(1 + b2)(1 + c2)(x2 + y2) + 2(a2b2c2 + 2abct(bc + ca + ab) + 2t(a2b + a2c + b2c + b2a + c2a + c2b) + 2t(a + b + c) – a2 – b2 – c2 – 2)x + 2(2a2b2c2t – abc(bc + ca + ab) – 2t(a2 + b2 + c2) – (a2b + a2c + b2c + b2a + c2a + c2b) – (a + b + c + 4t))y = 0. (2.8) This circle clearly passes through J and it may be verified by substitution that it also passes through H. Equation (2.8) therefore represents the Hagge circle, centre Q and passing through J and H. Fig. 1 has now been completely analysed and from here the results stated in the introduction follow by synthetic arguments and other known results. 3. Results
4
The angle arguments given below are appropriate to the figure at the start of the article and as such are diagram dependent. However, they are easily modified for other possible figures, or may be written in a diagram independent way by working mod π. Theorem 4.1 Angle JLB = Angle JYB = Angle JYH = 180o – Angle JZH = 180o – Angle JZC = Angle JLC. This shows that L lies on BC. Similarly, M and N lie on CA and AB respectively. Theorem 4.2 Now angle JNL = Angle JBL = = 180o – Angle JBC = Angle JAC = Angle JAM = 180o – Angle JNM and so LMN is a straight line. Theorem 4.3 Circles JBY and JCZ have centres Bo and Co respectively. The common chord JL is perpendicular to the line of centres BoCo and bisected by it. L is therefore the reflection of J in the side BoCo of triangle AoBoCo. The same holds for M and N with respect to the sides CoAo and AoBo respectively. LMN is therefore the Double Wallace-Simson line of J with respect to triangle AoBoCo. But, by construction, Q is the orthocentre of this triangle and therefore lies on LMN. Since LMN is drawn making right angles with the sides of a triangle which is similar and at a fixed angle to the sides of ABC, it is clear that these lines make equal angles with the sides of ABC. It is interesting to observe that these generalized Wallace-Simson lines (sometimes called oblique Wallace-Simson lines) exist only because Double Wallace-Simson lines exist for a smaller similar triangle. The original proof seems to have been by Carnot, the best reference I could find being in Exercises by Frére Gabriel-Marie []. References 1. F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73. 2. Frère Gabriel-Marie (listed as simply F. G.-M.) Exercices de géométrie, comprenant l'exposé des méthodes géométriques et 2000 questions résolues, 6ème édition. Paris: J. Gabay, 1991. 1302.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
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ARTICLE 8 A Singular Miquel Configuration and the Miquel Direct Similarity Christopher J Bradley
A
P D V R
S E F
B Q
U
Fig. 1
1
C
1. Introduction A Miquel configuration is one in which points P, Q, R are chosen on the sides AB, BC, CA respectively (but not at the vertices A, B, C) and the circles ARP, BPQ and CQR are drawn, illustrating that these three circles always pass through a common point S. The configuration we have called a singular Miquel configuration is that once P has been chosen arbitrarily (but not at B or C), then Q and R are chosen so that Q lies on circle CAP and R lies on the circle BCP. We prove in this paper that in these circumstances ASQ and BSR are straight lines and that the centres U, V, D, E, F of the five circles BCP, CAP, ARP, BPQ, CQR respectively do themselves lie on a circle passing through P. The Figure shows a case in which P lies between A and B, but the results still holds when P is external and the algebra we present is independent of the position of P on AB. We also show that triangle DEF is directly similar to ABC and as this is true in any Miquel configuration we give a proof when P, Q and R are arbitrary. In the analysis of the singular Miquel configuration we use areal co-ordinates in which ABC is the reference triangle. For the study of the direct similarity we use Cartesian co-ordinates. 2. Finding Q and R from P Suppose P is chosen to have co-ordinates (n, (1 – n), 0), n ≠ 0, 1. The general equation of a circle in areal co-ordinates is (2.1) Here u, v, w are determined by substituting the co-ordinates of three points lying on the circle. Using this method repeatedly we find the equations of CAP and BCP are respectively are (2.2) and (2.3) The co-ordinates of Q are therefore (0, a2 – c2n, c2n) and the co-ordinates of R are (b2 + c2(n – 1), 0, c2(1 – n)). The point S we define to be AQ^BR and has co-ordinates (n(b2 + c2(n – 1)), (1 – n)(a2 – c2n), c2n(1 – n)). 3. The three Miquel circles The circles ARP. BPQ have equations (3.1) (3.2) and it may be checked that both these circles pass through S. Finally, the circle CQR has equation 2
(3.3) and, of course, it too passes through S. 4. The centres of the 5 circles If a circle is expressed in the form of equation (2.1), then, see Bradley [1], its centre has coordinates x, y, z), where
(4.1) . Using Equation (4.1) and values of u, v, w from the working in Section 2 we find the coordinates of the centre of circle CAP are
(4.2)
and the co-ordinates of the centre of circle BCP are
(4.3)
The co-ordinates of the centre of circle ARP are
(4.4)
The co-ordinates of the centre of circle BPQ are
(4.5)
Finally the co-ordinates of the centre of the circle CQR are 3
(4.6)
The last three of these points was used to compute the circumcircle of triangle DEF and it was then checked that it does actually contain U, V and P. The circumcircle has a cumbersome equation which is
(4.7)
4
A
A' R
D
C'
F
S P E Q
B
C
B'
Figure 2
5. The direct similarity when P, Q, R are arbitrary For the more general Miquel configuration, when P, Q, R are chosen as arbitrary points on AB, BC, CA respectively, it is easier to use Cartesian co-ordinates. An economically chosen set of parameters involve co-ordinates (0, 0) for A, (2, 2v) for B and (2, 2w) for C, where without loss of generality we take w > v. Points P, Q, R on AB, BC, CA respectively are now assigned coordinates (k, kv), (2, 2u), (h, hw) respectively, where u ≠ v, w; h, k ≠ 0,2. See Figure 2. The equations of the circles BPQ, CQR, ARP may be obtained using standard methods involving 4 x 4 determinants and are respectively x2 + y2 + (2(uv – 1) – k(v2 + 1))x – 2(u + v)y + 2k(v2 + 1) = 0, (5.1) 2 2 2 2 x + y + (2(uw – 1) – h(w + 1))x – 2(u + w)y + 2h(w + 1) = 0, (5.2) 2 2 2 2 2 2 (v – w)(x + y ) + (kw(v + 1) – hv(w + 1))x + (h(w + 1) – k(v + 1))y = 0. (5.3)
5
It may be checked that these three circles have a common point S. We do not record its coordinates as they are complicated and are not needed in what follows. The co-ordinates of their centres can now be determined and are D (centre of ARP): {1/(2(v – w))}( hv(w2 + 1) – kw(v2 + 1), k(v2 + 1) – h(w2 + 1)), E (centre of BPQ): (½(k(v2 + 1) – 2(uv – 1)), u + v), F (centre of CQR) : (½(hw2 + 1) – 2(uw – 1)), u + w). Calculations may now be carried out to show that DE/EF = AB/BC = {√(v2 + 1)}/(w – v), and DF/DE = AC/AB = √{(w2 + 1)/(v2 + 1)} and this establishes the similarity between triangle XYZ and ABC. It may be shown that triangle ABC may be moved by appropriate rotation to A'B'C' and enlargement (reduction) about S as centre of direct similarity to coincide with triangle DEF. Flat 4, Terrill Court, 12-14 Apsley Road BRISTOL BS8 2SP
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Article 9 Miquel circles and Cevian lines Christopher J Bradley
A M Q W U
N
P V
C B
L
Figure 1
1
1.
Introduction
Given a triangle ABC it is well-known that if you choose L, M, N on the sides BC, CA, AB respectively (but not at the vertices), then circles AMN, BNL, CLM have a common point Q commonly known as the Miquel point [1, 2]. An interesting question is whether there exist any further properties when L, M, N are the feet of Cevian lines through a point P. We answer the question in the affirmative and establish the theorem that both P and Q lie on a circle UVW, where U is the intersection of AP with circle AMN, V is the intersection of BP with circle BNL and W is the intersection of CP with circle CLM. See Figure 1. We also have a second result which also only holds when L, M, N are the feet of Cevians and this is that circles AQL, BQM, CQN have a common point R. See Figure 2. The proofs given below involve areal co-ordinates. The definition of areal co-ordinates and how to use them may be found in Bradley [3].
A
M N
R Q P
B
C L
Figure 2
2
2. The Miquel circles Let P have co-ordinates (l, m, n) so that the co-ordinates of L, M, N are respectively L(0, m, n), M(l, 0, n), N(l, m, 0). The equation of any circle is (2.1) For circle AMN, we insert the co-ordinates of A, M, N in Equation (2.1) and find three equations for u, v, w, which give
We now substitute back in Equation (2.1)
to get the equation of AMN, which is (2.2) The equations of BMN, CNL may now be written down by cyclic change of x, y, z and a, b, c and l, m, n. 3.
The Miquel Point Q
It may now be checked that these three circles have a common point Q with co-ordinates (x, y, z), where
(3.1)
4. The points U, V, W The point U is the intersection of AP and circle AMN. AP has equation ny = mz and AMN has Equation (2.2). It follows that U has co-ordinates (x, y, z), where (4.1) The co-ordinates of V, W now follow by cyclic change of x, y, z and a, b, c and l, m, n. 5.
The circle UVWPQ
The co-ordinates of U, V, W are now put into an equation of the form (2.1). The result is three equations to determine u, v, w. These are now substituted back and the result is the equation of circle UVW, which is 3
(5.1) It may now be verified that both P and Q lie on this circle. 6.
The circles AQL, BQM, CQN
Starting from Equation (2.1) we may insert the co-ordinates of A, Q, and L to obtain three equations for u, v, w. Substituting these values back into the equation and simplifying we obtain the equation of circle AQL, which is
(6.1) The equations of the circles BQM, CQN may now be written down by cyclic change of x, y, z and a, b, c and l, m, n. Unfortunately the algebra computer package DERIVE was not powerful enough to deduce the co-ordinates of the point of intersection of these three circles, other than Q. 7.
The centres of the three circles
As the three circles are known to have the common point Q, then they will have a second common point R if, and only if, their centres lie on a line. This is because they are then bound to form part of a coaxal system of circles. We refer again to Bradley [3] for how to obtain the centre of a circle. What has to happen is that the polar of the centre is the line at infinity x + y + z = 0. This means that the centre of the conic with equation (7.1) must have co-ordinates X : Y : Z, where (7.2) Using Equation (6.1) we find the centre of circle AQL ha co-ordinates (X, Y, Z), where
(7.3)
4
The centres of circles BQM and CQN may now be written down by cyclic change of X, Y, Z and a, b, c and l, m, n. A 3 x 3 determinantal test with the co-ordinates of the centres of these three circles results in a zero value for the determinant, implying that the three centres are collinear, and hence that there must be a second point of intersection R. References
1. Miquel, A. "Mémoire de Géométrie." Journal de mathématiques pures et appliquées de Liouville 1, 485-487, 1838. 2. Honsberger, R. Episodes in Nineteenth and Twentieth Century Euclidean Geometry. Washington, DC: Math. Assoc. Amer., p. 81, 1995. 3. Bradley, C. J., The Algebra of Geometry, Highperception, Bath (2007).
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 10 A New Construction to find any Circle through a Given Point Christopher J Bradley
A
G P O W'
U'
W
U
K
F
Q
D
D'
V' V
B
C
E
Figure
1.
Introduction
In 1907 Hagge wrote an article [1] in which he provided a construction that always provides a circle, now called a Hagge circle, through the orthocentre H of a triangle ABC. In the construction there is a key point D such that lines ADE, BDF, CDG meet the circumcircle at E, F, G respectively. And then E, F, G are reflected in the sides BC, CA, AB respectively to provide three points U, V, W that are concyclic with H. When you vary 1
D you get a different circle through H. Later Peiser [2] showed that the isogonal conjugate of D, when rotated by 180o about the nine-point centre coincides with the centre of the Hagge circle. Furthermore triangles EFG and UVW are indirectly similar by a spiral similarity through the point D, as shown by the work of Speckman [3]. In the similarity triangle ABC is similar to a triangle XYZ inscribed in the Hagge circle, where triangle XYZ is in perspective to triangle UVW with vertex of perspective D. The disadvantage of the construction is the curious connection between the centre of the Hagge circle and the point D that generates it. The advantage is that because of the indirect similarity many interesting results, applications of the work by Speckman [3], exist for the configuration. It has long been an objective that we should be able to provide a similar construction to provide special circles through any point P (not lying on the sides of ABC) but that has remained elusive until now. At the expense of the similarity we now have a method, which has the advantage that one can predetermine immediately the radius and centre of the resulting circle through P. The method is as follows: If a special circle through P is required, then draw a segment POQ, where O is the circumcentre of triangle ABC and O is the midpoint of PQ. Then choose a point D so that OPKD is a parallelogram, where K is the desired centre of the circle and PK = OD is its desired radius. Then draw lines ADE, BDF, CDG where E, F, G lie on the circumcircle. Next construct the points U, V, W such that they are the fourth points of the parallelograms AQEU, BQFV, CQGW respectively. The key result is that circle UVW passes through P and has centre K and radius OD. A second result is that if the diagonals of these three parallelograms meet at U', V', W' respectively, then circle U'V'W' is a circle passing through O and D, with OD as diameter, and with half the radius of the special circle through P. In fact QUU', QVV', QWW' are straight lines, so that the circles and the disposition of the points on them are directly similar by an enlargement factor two through Q (showing incidentally that one can construct similar circles through any point on the line QOP by choosing points on the diagonals QU, QV, QW suitably). All these result are illustrated in the figure above. In the following sections we prove these results using Cartesian co-ordinates, taking the circumcircle to have radius 1 and centre at O. The working is technically quite elaborate, but the final results are beguilingly simple. 2.
Position of the special circle and the points U, V, W
Take P to have co-ordinates (– k, 0), so that Q being the 180o rotation of P about O has coordinates (k, 0). Let D have co-ordinates (m, n). What we shall establish is that the radius of the special circle UVWP is √(m2 + n2) and that its centre K has co-ordinates (m – k, n). Let A have co-ordinates (2a/(1 + a2), (1 – a2)/(1 + a2)), and let B, C have corresponding parameters b, c. The equation of the line AD is {(n + 1)a2 + (n – 1)}x – {(1 + a2)m – 2a}y + (1 – a2) m – 2an = 0. 2
(2.1)
This line meets the circumcircle with equation x2 + y2 = 1 at the point E with co-ordinates (x, y), where x = 2{m(n + 1)a2 – (1 + m2 – n2)a + m(1 – n)}/{(m2 + (n + 1)2)a2 – 4ma + m2 + (1 – n)2}, y = {(n + 1)2 – m2)a2 – 4mna + m2 – (1 – n)2}/{(m2 + (n + 1)2)a2 – 4ma + m2 + (1 – n)2}. (2.2) Having obtained E the co-ordinates of the displacement QE may be obtained. Bearing in mind that AU = QE, the co-ordinates of U may now be found and they are (x, y), where x = (1/s){– (k(m2 + (n + 1)2) – 2m(n + 1))a4 + 4(km + n(n + 1))a3 – 2(k(m2 + n2 + 1) + 2m)a2 + 4(km + n(n – 1))a – k(m2 + n2 – 2n + 1) – 2m(n – 1)}, y = (1/s) {– 2(m2a4 + 2m(n – 1)a3 – 4na2 + 2m(n + 1)a – m2)}, (2.3) and s = (1 + a2) {(m2 + (n + 1)2)a2 – 4ma + m2 + (n – 1)2}. (2.4) The co-ordinates of V and W may now be obtained by replacing the letter a in Equations (2.2) – (2.4) by letters b and c respectively. 3.
The circle UVWP
The equation of the circle with centre (– g, – f) and radius √(g2 + f2 – t) is x2 + y2 + 2gx + 2fy + t = 0. (3.1) Inserting the co-ordinates of U, V, W we obtain three equations for f, g, t. When these are substituted back in Equation (2.5), we obtain the equation of circle UVW, which is x2 + y2 + 2x(k – m) – 2mny + k(k – 2m) = 0. (3.2) It can now be seen that circle UVW passes through P(– k, 0), has centre K(m – k, n) and radius √(m2 + n2), as required to be proved. 4.
The points U'V'W' and the circle U'V'W'DO
Having obtained the co-ordinates of U, it is straightforward to obtain the co-ordinates of U', the midpoint of QU. Its co-ordinates (x, y) are independent of K, since U' is also the midpoint of AE. They are given by x = (1/s){(n(1 + a2) – (1 – a2))(2an – m(1 – a2))}, y = (1/s){(2a – m(1 + a2))(2an – m(1 – a2))}, (4.1) where s is given by Equation (2.4). The co-ordinates of V' and W' may be written down from Equation (4.1) by replacing the letter a by letters b and c respectively. The equation of circle U'V'W' may now be obtained and is x2 + y2 – mx – ny = 0. (4.2) This circle clearly passes through both O and D and has centre (½m, ½n). In fact OD is a diameter and its radius is ½√(m2 + n2).
5.
Another circle through O
Finally by letting P, O, Q coincide at O and forming the parallelograms AOEU'', BOFV'', COGW'' the circle U''V''W'' passes through O and has centre D.
3
References 1. K. Hagge, Zeitschrift für Math. Unterricht, 38 (1907) 257-269. (German) 2. A.M Peiser, Amer. Math. Monthly, 49 (1942) 524-527. 3. H. A. W. Speckman, Perspectief Gelegen, Nieuw Archief, (2) 6 (1905) 179 – 188. (Dutch)
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
4
Article 11 P1
Eight circles through the Orthocentre Christopher J Bradley D
W2 W1
V3
N' A
N W' U
G
V V'
E
B
P2
M
H
M' Q
U'
W
C
V1
L' L
P3 U2 F
U3
Figure
1.
Introduction
An appreciation of the Figure (drawn with CABRI II plus) requires a description of how it was drawn and the order in which the various constructions were made. Start with a triangle ABC and its centroid G. AG, BG, CG meet the circumcircle Γ of ABC at points L, M, N respectively. 1
Tangents at L, M, N to Γ are then drawn with those at M and N meeting at D. E and F are similarly defined. Γ then becomes an incircle of triangle DEF. Next DA, EB, FC are drawn and these turn out to meet at a point Q and when extended meet Γ again at L', M', N' respectively. Lines AL, BN', CM' now prove to be concurrent at P1; BM, CL', AN' prove to be concurrent at P2; and CN, AM', BL' prove to be concurrent at P3. Furthermore it transpires that P1, P2, P3 are collinear lying on the polar of Q with respect to Γ. Twelve more points are now obtained as reflections of L, M, N, L', M', N' in the sides of ABC. These are U, V, W the reflections of L, M, N respectively in the sides BC, CA, AB; U', V', W' are the reflections of L', M', N' respectively in the sides BC, CA, AB; U2, U3 are the reflections of M', N' respectively in the side BC; V3, V1 are the reflections of N', L' respectively in the side CA; and finally W1, W2 are the reflections of L', M' respectively in the side AB. Now five circles can be drawn, all of which pass through the orthocentre H, UVW is the Hagge circle of G, U'V'W' is the Hagge circle of Q, UV3W2 is the Hagge circle of P1, U3VW1 is the Hagge circle of P2 and U2V1W is the Hagge circle of P3. For the original account of Hagge circles see Hagge [1], for a modern account see Bradley and Smith [2]. Three more circles may now be drawn, circles BHC, CHA, AHB, and these circles remarkably contain the points U, U', U2, U3 and V, V', V3, V1 and W, W', W1, W2 respectively. It is also the case that lines V1W1, W2U2 and U3V3 all pass through H. The analysis that follows in later sections proves all the above results using areal co-ordinates with ABC as triangle of reference. For an account of these co-ordinates and how to use them, see Bradley [3]. It must now be confessed that the point G was used in the construction and in the analysis that follows in order to provide an analysis that is as easy as possible, the areal coordinates of G being simply (1, 1, 1). The truly remarkable thing is that all the concurrences, collinearities, and dispositions of points on circles holds not just for the starting point G, but for a any internal point of ABC taken as starting point (except H itself and except the symmedian point K when in both cases the figure degenerates). The problem, of course, is that if one starts with P(l, m, n) the analysis becomes technically very difficult (because of the three extra variables) even with a good algebra computer package, but not only that, it becomes virtually unprintable. No excuses are therefore made for having taken the easiest option. 2.
The points L, M, N, D, E, F
The equation of the line AG is y = z and this meets the circumcircle Γ with equation (2.1) at the point L with co-ordinates L(– a2, b2 + c2, b2 + c2). M and N follow by cyclic change and have co-ordinates M(c2 + a2, – b2, c2 + a2) and N(a2 + b2, a2 + b2, – c2). The equation of the tangent to Γ at any point (d, e, f) is (2.2) 2
It follows that the equation of the tangent at L is (2.3)
The tangents at M and N are respectively (2.4) and (2.5) The tangents at M and N meet where both equations (2.4) and (2.5) hold. This is at the point D with co-ordinates D(– (a4 + a2(b2 + c2) + 2b2c2), b2(a2 + b2 – c2), c2(c2 + a2 – b2)). The coordinates of E and F follow by cyclic change and are E(a2(a2 + b2 – c2), – (b4 + b2(c2 + a2) + 2c2a2, c2(b2 + c2 – a2)) and F(a2(c2 + a2 – b2), b2(b2 + c2 – a2), – (c4 + c2(a2 + b2) + 2a2b2)). 3.
The points Q, L', M', N', P1, P2, P3
The critical result that makes the configuration possess so many interesting details is the concurrence of the lines DA, EB, FC at the point Q. This is the point that links L, M, N with the points L', M', N'. Without that link the figure would not contain enough circles to be of significance. The equations of the lines DA, EB, FC are (3.1) (3.2) (3.3) respectively. These lines are concurrent at the point Q with co-ordinates (x, y, z), where Note that the isogonal of Q is the isotomic of H. Now the line DA with Equation (3.1), when extended, meets Γ, with Equation (2.1), at the point L', which therefore has co-ordinates L'(– 1, 2b2/(c2 + a2 – b2), 2c2/(a2 + b2 – c2)). The co-ordinates of M' and N' follow by cyclic change and are M'(2a2/(b2 + c2 – a2), – 1, 2c2/(a2 + b2 – c2)) and N'(2a2/(b2 + c2 – a2), 2b2/(c2 + a2 – b2), – 1). The polar of a point (d, e, f) with respect to Γ has the same form as Equation (2.2) and accordingly the equation of the polar of Q with respect to Γ is (3.4) Next it emerges that the three lines AL, BN', CM' are concurrent at the point P1 with co-ordinates P1(2a2, a2 – b2 – c2, a2 – b2 – c2), and the three lines BM, CL', AN' are concurrent at P2 with coordinates P2(b2 – c2 – a2, 2b2, b2 – c2 – a2), and the three lines CN, AM', BL' are concurrent at P3 3
with co-ordinates P3(c2 – a2 – b2, c2 – a2 – b2, 2c2). Finally P1, P2, P3 are collinear and lie on the line with Equation (3.4), which is the polar of Q with respect to Γ. Note that we have now shown that triangle ABC is in perspective with each of the five triangles LMN, L'M'N', LN'M', N'ML', M'L'N, the five vertices of perspective being G, Q, P1, P2, P3 respectively. 4. Definition of the points U, V, W, U', V', W', U2, U3, V3, V1, W1, W2 The points U, V, W are the reflections of L, M, N respectively in the sides BC, CA, AB. The points U', V', W' are the reflections of the points L', M', N' respectively in the sides BC, CA, AB. The points U2, U3 are the reflections of M', N' respectively in the side BC. The points V3, V1 are the reflections of the points N', L' respectively in the side CA and the points W1, W2 are the reflections of the points L', M' in the side AB. To go through all the details of obtaining the coordinates of these points would be both dull and unnecessary. Instead we give expressions for the reflection of a general point with co-ordinates (d, e, f) in the sides BC, CA, AB and then quote the co-ordinates of those points that are necessary to establish the features illustrated in the figure and explained in section 1. The treatment of perpendicularity when using areal co-ordinates is tiresome. Details of how to proceed are given in Bradley [3, 4]. Following the plan explained above we find the reflection of (d, e, f) in the side BC has co-ordinates (– d, e + d(a2 + b2 – c2)/a2, f + d(c2 + a2 – b2)/a2), the reflection of (d, e, f) in the side CA has co-ordinates (d + e(a2 + b2 – c2)/b2, – e, f + e(b2 + c2 – a2)/b2) and the reflection of (d, e, f) in the side AB has co-ordinates (d + f(c2 + a2 – b2)/c2, e + f(b2 + c2 – a2)/c2, – f). 5.
The five Hagge circles
Using the results of Section 4 we find the co-ordinates of U to be (a2, 2c2 – a2, 2b2 – a2), the coordinates of V to be (2c2 – b2, b2, 2a2 – b2) and the co-ordinates of W to be (2b2 – c2, 2a2 – c2, c2). The general equation of a circle in areal co-ordinates is (5.1) We put the co-ordinates of U, V, W into Equation (5.1) and obtain three equations for u, v, w. These are then reinstated in Equation (5.1) and simplified to obtain the Hagge circle of G, whose equation is
(5.2) It may be checked that this circle passes through the orthocentre H with co-ordinates (1/(b + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)). 2
4
We now repeat this exercise to find the co-ordinates of the points U', V', W' and hence determine the equation of the Hagge circle of Q. The co-ordinates of U' turn out to be (1, – (a4 – 2a2b2 – (b2 – c2)2)/{a2(c2+ a2 – b2)}, – (a4 – 2c2a2 – (b2 – c2)2)/{a2(a2 + b2 – c2)}). The co-ordinates of V' and W' may be written down by cyclic change. Using the method described above we find the Hagge circle of Q to be
(5.3) Again this circle contains H. The three other Hagge circles are UV3W2, VW1U3 and WU2V1 and derive from the points P1, P2, P3 respectively. We do not give the equations of these circles. 6.
Three lines and three more circles
The reflections of the point L', in the sides BC, CA, AB, are the points U', V1, W1 respectively. These are known to lie on a line through the orthocentre H, sometimes known as the Double – Simson line of L'. Similarly the points V', W2, U2, H are collinear as are W', U3, V3, H. The other three circles in the figure (besides Γ) are the circles BHC, CHA and AHB. These are well-known circles. The equation of CHA, for example, is – (6.1) The interesting result is that this circle passes through all the points V, V', V1 and V3. The coordinates of V and V' have already been given, those of V1 are (x, y, z), where
(6.2) and the co-ordinates of V3 are (x, y, z), where ,
(6.3)
Similarly circle BHC contains U, U', U2, U3 and circle AHB contains the points W, W', W1, W2. 7.
When the starting point is H or K
When the starting point is the orthocentre H, then the points U, V, W coincide with H and one of the Hagge circles becomes a point. Circles such as UV2W3 are ill-defined, though, of course, the circle HV2W3 can be drawn. When the starting point is the symmedian point K, then, as D, A, L are collinear, L, L' coincide, as do M, M' and N, N'. There are only four Hagge circles as Q coincides with K. It is interesting however, that triangle ABC is in perspective with the four triangles LMN, LNM, NML and 5
MLN. The latter three provide what is called a triple reverse perspective and triangles ABC and LMN become triangles that define the Brocard porism. The line P1P2P3 is now the polar of K and it contains three further interesting points AB^DE, BC^EF and CA^FD.
References 1. K.Hagge, Zeitschrift Für Math. Unterricht, 38, pp257-269 (1907). 2. C. J Bradley & G.C.Smith, Math. Gaz. pp202 -207 July 2007. 3. C. J. Bradley, Challenges in Geometry, Oxford (2005). 4. C. J .Bradley, The Algebra of Geometry, Highperception, Bath (2007).
Flat 4, Terrill Court, 12/14 Apsley Road Clifton BRISTOL BS8 2SP
6
Article 12 Circles concentric with the Circumcircle Christopher J Bradley
C' A Z E
F
V U Q
B'
O
P X
Y B
C W
D A'
Figure 1. Introduction Let ABC be a triangle with circumcircle S, centre O, and suppose P is a point not on the sides of ABC nor on S. In this short article we show that the circle, centre O, through P contains seven
1
points U, V, W, X, Y, Z, Q with special properties. The figure above illustrates these properties which are a result of the following three theorems. Theorem 1 Let AO, BO, CO meet S at D, E, F respectively. Let AP, BP, CP meet S at A', B', C' respectively. Draw lines l, m, n through P parallel to A'D, B'E, C'F respectively. Draw the line through D perpendicular to l to meet l at U. Define V, W similarly using E, m and F, n respectively. Then U, V, W lie on the circle Σ through P, centre O. Theorem 2 Draw the diameters UOX, VOY, WOZ of Σ then X, Y, Z lie on AP, BP, CP respectively. Theorem 3 AU, BV, CW are concurrent at a point Q lying on Σ. In the proofs that follow we use Cartesian co-ordinates. 2. Proof of Theorem 1 Take O to be the origin, S to be the unit circle, with equation x2 + y2 = 1, and A to have coordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)), with B and C having parameters b, c respectively. Let P have co-ordinates (c, 0), c ≠ 0, 1. The equation of AP is therefore 2ax + y(a2(c + 1) + c – 1) = 2ac.
(2.1)
This line meets S again at the point A' with co-ordinates (x, y), where x = {a2(c + 1)2 – (c – 1)2}/{a2(c + 1)2 + (c – 1)2}, y = {2a(c2 – 1)}/ {a2(c + 1)2 + (c – 1)2}.
(2.2) (2.3)
Since AOD is a diameter, the co-ordinates of D are the negatives of those of A and thus the equation of A'D is x(a2(c + 1) + (c – 1)) – 2ay = a2( c + 1) – (c – 1). (2.4) The line l is parallel to this through P and therefore has equation x(a2(c + 1) + (c – 1)) – 2ay = c(a2(c + 1) + (c – 1)).
(2.5)
The equation of the line through D perpendicular to A'D is 2ax + y(a2( c + 1) + (c – 1)) = – 2ac.
(2.6)
2
The lines with equations (2.5) and (2.6) meet at the point U with co-ordinates (x, y), where x = {c(a4(c + 1)2 +2a2(c2 – 3) + (c – 1)2)}/{a4(c + 1)2 + 2a2(c2 + 1) + (c – 1)2}, (2.7) 2 4 2 2 2 2 y = – {4ac(a ( c + 1) + (c – 1)}/ {a (c + 1) + 2a (c + 1) + (c – 1) }. (2.8) The sum of the squares of these co-ordinates is c2, so U lies on the circle Σ, centre O, radius c, which passes through P. Similarly V, W lie on Σ. 3. Proof of Theorems 2 and 3 Since UOX is a diameter of Σ the co-ordiantes of X are the negatives of those of U and it is soon checked that X lies on AP with Equation (2.1). The line DU with equation (2.6) meets Σ again at the point Q with co-ordiantes (– c, 0) and since its co-ordinates are independent of Q also lies on EV and FW. It also lies on the diameter PO.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
3
Article 13 Ex-points and eight intersecting circles Christopher J Bradley
H'
F
P3
L'
W'
A V' E
N M U
Q
V P
O
J
H W
M' R
B
C N'
L
To T
S
U' P1
D
Figure
1.
Introduction
1
P2
In this article we use areal co-ordinates throughout. See Bradley [1, 2 ] for a review of how to use them. Incentres and ex-centres are well-known and if (a, b, c) are the co-ordinates of the incentre, then the three ex-centres I1, I2, I3 have co-ordinates (– a, b, c), (a, – b, c), (a, b, – c) respectively. Less well-known, perhaps, are the ex-symmedian points D, E, F, which are the intersections of the tangents at the vertices of the circumcircle, D being the intersection of the tangents at B and C, with E, F similarly defined. These points can be seen in the Figure. Their co-ordinates are (– a2, b2, c2), (a2, – b2, c2), (a2, b2 – c2) respectively. It is also true that if P (l, m, n) is any point internal to a triangle ABC, then there are three associated ex-points P1(– l, m, n), P2(l, – m, n), P3(l, m, – n). We show how these points may be constructed using compass and straight edge only. It turns out that the construction is a development of the construction of the Hagge circle [3] of the point P, so we are able to draw four Hagge circles, one for each of P, P1, P2, P3, all of which, of course, pass through the orthocentre H of triangle ABC. As a bonus there are in addition four other circles and the CABRI II geometry computer software package indicates that these other four circles also have a common point, labelled H' in the Figure. The additional four circles have equations that are considerably more complicated than those of the four Hagge circles, so we have been unable to find the co-ordinates of H'. However, it may be proved by inversion that these four circles do indeed have a common point. This proof has been supplied by David Monk [4] and I am extremely grateful to him for providing the proof. We now describe the construction of the ex-points, starting from ABC and P. First draw the tangents at A, B, C to the circumcircle to provide D, E, F. Next draw the lines AP, BP, CP to meet the circumcircle at L, M, N respectively. Now draw DL, EM, FN to meet the circumcircle again at L', M', N' respectively. Note that these three lines are concurrent at a point Q, which we call the generating point. Now AL, BM', CN' are found to be concurrent at the point P 1, AL', BM, CN' are concurrent at P2, and AL', BM', CN are concurrent at P3. Section 2 provides the algebraic proof of these remarks. Q is called the generating point, because rather than start with P, one can start with Q, which must be internal to triangle DEF, and then reverse the construction, thus generating all 4 points P, P1, P2, P3. We defer the description of the construction of the 8 circles until later in the text. 2.
The ex-points P1, P2, P3 and the generating point Q
For those unfamiliar with areal co-ordinates for points on the circumcircle Γ we give a brief account of what is required. The co-ordinates of a general point on Γ are (– a2t(1 – t), b2(1 – t), c2t), where t is a parameter that can take any value (including infinity). Thus the points A(1, 0, 0), B(0, 1, 0), C(0, 0,1) have parameters ∞, 0, 1 respectively and the equation of the circumcircle is . (2.1) For checking purposes it is useful to have the equation of the chord of Γ joining points with parameters t and s. This is 2
(2.2) Starting with P(l, m, n) we find the equation of AP to be ny = mz. This meets Γ, with Equation (2.1), at the point L with co-ordinates L(– a2mn/(b2n + c2m), m, n). Similarly BP, CP meet Γ at M, N respectively with co-ordinates M(l, – b2nl/(c2l + a2n), n) and N(l, m, – c2lm/(a2m + b2l). The tangents to Γ with Equation (2.1) at A, B, C are respectively b2z + c2y = 0, c2x + a2y = 0, a2y + b2x = 0.The point D, that is the intersection of the tangents at B and C, therefore has coordinates D(– a2, b2, c2) and E, F respectively have co-ordinates E(a2, – b2, c2), F(a2, b2, – c2). It can now be seen that D, E, F are the ex-symmedians, as the symmedian point K has co-ordinates (a2, b2, c2). The line DL has equation (2.3) This meets Γ again at the point L' with co-ordinates L'(a2mn/(b2n – c2m), – m, n) Note that this differs from L simply by a change of sign of m (or n). We can now write down, by cyclic change, the co-ordinates of M' and N', which are M'(l, b2nl/(c2l – a2n), – n) and N'(– l, m, c2lm/(a2m – b2l)). Now EM has equation that may be derived from that of DL by cyclic change and then the generating point Q is determined as DL^EM. The result is that Q has co-ordinates It may now be checked that Q also lies on the line FN. The equations of the lines AL', BM', CN' can now be determined and they are respectively (2.4) We now have enough to determine the co-ordinates of the points P1 = AL^ BM'^CN', P2 = AL'^BM^CN' and P3 = AL'^BM'^CN, and, as you will have guessed, these are the ex-points of P with co-ordinates P1(– l, m, n), P2(l, – m, n), P3(l, m, – n). It may be checked that EP2 and FP3 intersect at a point R on BC, with S, T on CA, AB respectively defined similarly. In the Figure we have chosen P to be the isogonal conjugate of the nine-point centre. This point generates the Hagge circle with OH as diameter. 3.
A 9 point hyperbola
Using the standard 6 x 6 determinantal method to obtain the conic through P, its 3 ex-points and K the symmedian point we find the equation of a hyperbola (3.1) 3
Its form immediately indicates that not only does this hyperbola pass through the points just mentioned, but it also passes through D, E, F the ex-symmedians. It may be checked that it also passes through the generating point Q. Thus, it passes through 9 key points of the figure. As P may be any point, we have the result, for example, that a 9 point hyperbola passes through the incentre, the ex-centres, the symmedian point, the ex-symmedians and the point Q, which in this case is the circumcentre O. It is a rectangular hyperbola, since the incentre and ex-centres form an orthocentroidal quartet. If P lies at O, then the hyperbola is also rectangular, since the orthocentre of triangle O1O2O3 lies on the hyperbola. 4.
The reflected points
The points in the Figure labelled U, V, W, U', V', W' are the reflections of the points L, M, N, L', M', N' in the sides BC, CA, AB, BC, CA, AB respectively. This is the same as in the construction of Hagge circles. The algebra for obtaining the co-ordinates of the reflected points is covered in detail in Bradley and Smith [3] and is not repeated here. We therefore state the coordinates of these points without proof and they are as follows: (4.1) (4.2) (4.3) (4.4) (4.5) (4.6)
5.
The 8 circles
There is, of course a circle through any 3 of these points, but there are 8 special circles found by choosing one of U, U', one of V, V' and one of W, W'. First we consider the circle UVW. This is the Hagge circle generated by P. The method is to write down the general equation of a circle, which is (5.1) Then we insert the co-ordinates of U, V, W in turn, to provide three equations for the real numbers u, v, w. Their values are then substituted back in Equation (5.1). The result is
(5.2) It may be checked that this circle passes through H, the orthocentre of ABC. The circle UV'W' has the equation derived from Equation (5.2) by changing the sign of l. It is therefore the Hagge circle generated by P1. Similarly changes of sign of m, n in Equation (5.2) produce the Hagge circles of P2, P3 respectively. 4
The other four circles U'V'W', U'VW, UV'W, UVW' have equations that are extremely lengthy and complicated, and the algebra computer package DERIVE was unable to solve the simultaneous quadratics of any pair of them to show that they have the common point, labelled H' in the figure. However, David Monk [4] has kindly provided the following proof: Invert the figure in H so that circles UVW, UV'W', U'VW', U'V'W become the lines uvw, uv'w', u'vw', u'v'w. The resulting figure is a complete quadrilateral, and thus the four triangles u'v'w', u'vw, uv'w, uvw' have circumcircles that possess a common point h'. Inverting back gives the result required. This proof does not tell us anything about the dependence of H' on the point P. All that we can say from drawings is that the locus of H' as P moves on a line is a curve that is not a conic.
References 1. C.J.Bradley, Challenges in Geometry, Oxford (2005); 2. C.J.Bradley, The Algebra of Geometry, Highperception, Bath (2007); 3. Christopher Bradley and Geoff C. Smith, On a construction of Hagge, Forum Geometricorum p231—247 (2006). 4. D. Monk (private communication).
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5
Article 14 Intersecting Circles having chords the sides of a Cyclic Quadrilateral Christopher J Bradley
F A B
L
S
O
K P
N
X E
T
H
G C M
D
Figure 1
1.
Introduction
1
We consider a cyclic quadrilateral in which the centre O of the circle ABCD is distinct from the intersection E of the diagonals AC and BD. In this configuration a number of results hold. (i) Circles AOB, COD, AED, BEC are concurrent at a point S; (ii) Circles AOD, BOC, AEB, CED are concurrent at a point T; (iii)O, S, E, T are concyclic; (iv) The centre of circle OSET is X, the midpoint of OE, and it follows that angle OSE and OTE are each 90o; (v) FXG, NXP, HXK, LXM are straight lines, where F, G, H, K, L, M, N, P are the centres of circles AOB, COD, AOD, BOC, AEB, CED, AED, BEC respectively; (vi) The following angles are each 90o: angles HXL, NXF, KXL, PXF, HXM, NXG, KXM and PXG and consequently angles HXN, FXL, GXM, KXP are equal. See Figure 1 for an illustration of these properties. Now let AB and CD meet at F, and let AD and BC meet at G. Then another set of properties hold. In properties (vii) to (xi) points S, T, E are as before but points F, G, H, K are different points. (vii) Circles FBC, FAD, GCD, GAB meet at a point H lying on FG; (viii) Circles FBD, GAC pass through S and circles FAC, GBD pass through T; (ix) Points O, E, H are collinear; (x) OS^ET = F and OT^SE = G; (xi) Points F, G, S, T are concyclic and the centre K of the circle FGST is the midpoint of FG. See Figure 2 for an illustration of these properties. In the following sections we prove these results, using Cartesian co-ordinates. 2.
Circles AOB and COD and the point S
We take the equation of ABCD to be x2 + y2 = 1 and the co-ordinates of A to be (2a/(1 + a2), (1 – a2)/(1 + a2)), with B, C, D having similar co-ordinates with parameters b, c, d respectively. The co-ordinates of O are, of course, (0, 0). Since it passes through O the equation of circle OAB is of the form (2.1) Inserting the co-ordinates of A and B in Equation (2.1) we find f and g and substituting back we obtain the equation of circle OAB to be (2.2)
2
The equation of OCD may be obtained from Equation (2.2) by writing c, d instead of a, b. The intersections of these two circles are the origin O and the point labelled S, whose coordinates are (x, y), where (2.3)
(2.4) and
.
(2.5) 3.
Circles AOD and BOC and the point T
The equations of circles AOD and BOC may be written down by altering the parameters in Equation (2.2) to a, d and to b, c respectively. The co-ordinates of T may be written down from Equations (2.3) to (2.5) by exchanging a and c (or b and d, but not both). 4. The point of intersection E of the diagonals AC and BD The equation of AC is (4.1) and the equation of BD is (4.2) These lines meet at E with co-ordinates (x, y), where (4.3) (4.4) and
5.
The circles AED, BEC, AEB, CED
The general equation of a circle is of the form (5.1) If we now insert the co-ordinates of A, E, D into Equation (5.1) we obtain three equations for f, g, k. When these are substituted back into Equation (2.1) we find the equation of AED, which is
3
. (5.2) The equations of circles BEC, AEB, CED may now be written down from Equation (5.2). Bearing in mind that the co-ordinates of E are invariant under the interchange of a and c and of b and d, it is now straightforward to write down the equations of circles BEC, AEB, CED. 6.
(i) and (ii) and the circle OSET
It may now be verified that circles AED and BEC pass through S and circles AEB and CED pass through T. The circle OES has an equation of the form (2.1) and inserting the co-ordinates of E, given in Equations (4.3) to (4.5) and the co-ordinates of S, given in Equations (2.3) to (2.5) we get two equations for f and g. When these values are reinserted in (2.1) we obtain the equation of OES, which is
(6.1) It may now be verified that T lies on this circle. 7.
Centres of circles and properties (iv), (v), (vi)
The centre of the circle with Equation (5.1) is, of course (– g, – f). It follows that the centre X of circle OSET has co-ordinates (1/k)(ac – bd, – ½ (7.1) where k= . (7.2) Noting Equations (4.3) to (4.5) we see that X is the midpoint of OE. This establishes property (iv).
The perpendicular bisector of AB passes through the centres of circles ABCD, AOB, AEB. It follows that points F, L, O are collinear. Similarly H, N, O and P, K, O and M, G, O are collinear, all four lines meeting at O. From Equation (2.2) the centre F of circle AOB has co-ordinates (½(a + b)/(1 + ab), ½(1 – ab)/((1 + ab)). Similarly the centre G of circle COD has co-ordinates (½(c + d)/(1 + cd), ½(1 – cd)/((1 + cd)). A 3 x3 determinantal test using the co-ordinates of X, F, G in the first two columns and 1 as the entry in each row of the third column gives a zero result. Hence F, X, G are collinear. Similarly H, X, K are collinear. 4
From Equation (5.2) the centre N of circle AED has co-ordinates (1/k)((a + d)(c – b), (c – b)(1 – ad)), where k = a(b(c – d) + cd + 1) – b(1 + cd) + c – d. The co-ordinates of the centre P of circle BEC may now be written down by exchanging b and d and also a and c. It may now be verified that N, X, P are collinear and similarly L, X, M are collinear. This now establishes property (v). We now have the co-ordinates of N, X and F and can therefore find the equations of the lines FX and NX. The equation of FX is (7.3) and the equation of NX is
(7.4) From these equations it is evident that the product of their gradients is – 1 and hence angle FXN = 90o. Similarly the angles LXH, PXF, KXL, GXN, MXH, MXK and GXP are all 90o. This completes the proof of property (vi). 8.
The points F and G and the line FG
The intersection of lines AB and CD is the diagonal point F and its co-ordinates are therefore (x, y), where (8.1)
(8.2)
5
A
O
B S
E
T
F C D
HK
G
Figure 2
where (8.3) The diagonal point G is the intersection of lines AD and BC and therefore has co-ordinates (x, y), where (8.4) (8.5) and (8.6) The equation of the line FG may now be obtained and is
(8.7)
9.
OT and SE pass through G and OS and TE pass through F 6
The equation of OT is (9.1) It may now be verified, using Equations (8.4) to (8 .6) that G lies on this line. The equation of SE is (9.2) It may be verified that G also lies on this line. Similarly F lies on OS and TE. This establishes property (x). 10.
Circles FBC and FAD, the point H, the line OEH and the circle FSTG
Following the usual procedure we obtain the equation of circle FBC, which is
(10.1) And the equation of circle FAD is
(10.2) The intersection of these circles, other than the point F, is the point we define to be the point H and its co-ordinates are (x, y), where
(10.3)
(10.4) where
.
(10.5)
It may now be verified that H lies on FG with Equation (8.7). The equations of circles GCD and GAB follow by carefully rearranging parameters in Equations (10.1) and (10.2) and then it may be checked that H lies also on these circles. This establishes property (vii). The line OE passes through H and as O is the orthocentre of triangle EFG it follows that OE cuts FG at right angles. We have already shown that angles OSE and OTE and that OS passes through F and OT passes through G. It follows that angles FSG and FTG are both right angles. Thus S, T, F, 7
G are concyclic and the centre K of circle FSTG is the midpoint of FG. We have now established properties (ix) and (xi). 11. Circles FBD and FAC pass through T and circles GBD and GAC pass through S From the co-ordinates of F, B, D we may now obtain the equation of circle FBD, which is
.
(11.1)
It may now be checked that T lies on this circle. Similarly T lies on circle FAC, S lies on circle GBD and also on circle GAC. This establishes property (viii). Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 15 Four Concurrent Euler Lines Christopher J Bradley 1. Introduction In this paper we consider a cyclic quadrilateral PQRS in which the diagonals PR and QS meet at a point E and we prove a number of results about the triangles PQE, QRE, RSE, SPE.
These are: 1. The Euler lines of the four triangles are concurrent at a point X;
1
2. If the circumcentres of the four triangles are denoted by O1, O2, O3, O4 respectively then O1O2O3O4 is a parallelogram; 3. If the orthocentres of the four triangles are denoted by H1, H2, H3, H4 respectively then H1H2H3H4 is a parallelogram; 4. The parallelograms O1O2O3O4 and H1H2H3H4 are inversely similar, with O1O2 parallel to H1H4 etc.; 5. O1H3, O3H1, O2 H4, O4H2 are concurrent at E. Results 2 and 3 are known results, see Bradley [1], but they are inevitably given explicit proof in the course of the calculations required to prove the other results. The results are illustrated in the figure. It is also the case that the analysis takes no account of the order of the vertices P, Q, R, S and is equally valid when P and R are adjacent vertices. This means that Results 1-5 are also true for the cyclic quadrilateral PQRS when E is replaced by F = PQ^RS or G =QR^SP, with the four triangles replaced in an obvious way. Analysis is carried out using Cartesian co-ordinates. Singular cases can occur when one or other of the key points recede to infinity. We assume in our analysis that all points arising are finite and distinct. Calculations were carried out using the algebra package DERIVE and have been subject to various consistency tests. 2. Preliminary results We consider a triangle ACE, with E(0, 0), A(a, b), C(c, d), (ad ≠ bc), and work out the coordinates of its circumcentre O, its orthocentre H and the equation of its Euler line OH. In subsequent sections we identify the vertices A, C successively as the pairs P, Q; Q, R; R, S; S, P. The equations of EA, EC, AC are immediate and are respectively ay = bx, (2.1) cy = dx (2.2) and x(b – d) + y(c – a) + ad – bc = 0 (2.3) The perpendicular bisectors of EA and EC are respectively 2(ax + by) = a2 + b2 and 2(cx + dy) = c2 + d2.
(2.4) (2.5)
These meet at the circumcentre O with co-ordinates (x, y), where x=
(2.6)
y=
.
(2.7)
It may be checked that this also lies on the perpendicular bisector of the line AC. The three altitudes have equations 2
y(b – d) = x(c – a), ax + by = ac + bd, cx + dy = ac + bd.
(2.8) (2.9) (2.10)
These meet at the orthocentre H with co-ordinates (x, y), where x=
–
(2.11)
y=
.
(2.12)
We may now calculate the equation of the line OH, the Euler line of triangle ACE, and the result is (3ac(a – c) + ad(2b – d) + bc(b – 2d))x + (3bd(b – d) + ad(a – 2c) + bc(2a – c))y – (a2 + b2 – c2 – d2)(ac + bd) = 0. (2.13) It may be checked that the centroid ⅓(a + c, b + d) lies on this line. 3. Co-ordinates of key points in the configuration The ease with which a calculation may be carried out depends critically on the choice of the coordinates of E, P, Q, R, S. As in Section 2 the point E is taken as origin. For the points P, Q, R, S we take QS to be the x-axis and PR to be inclined to it in a direction determined by the unit vector (
) and use the converse of the intersecting chord theorem to specify their co-
ordinates as P(km(1 – t2), 2kmt), Q(–mn(1 + t2), 0), R(– nl(1 – t2), – 2nlt), S(kl(1 + t2), 0). With these co-ordinates the equation of the circle PQRS is 2t(x2 + y2) + 2t(1 + t2)(mn – kl)x + (1 + t2){(nl – km)(1 + t2) + (kl – mn)(1 – t2)}y – 2klmnt(1 + t2)2 = 0. (3.1) With E as origin then the analysis of Section 2 may be applied to each of the triangles PQE, QRE, RSE, SPE in turn by varying the values of a, b, c, d. Result 1 The four Euler lines of these triangles are concurrent at the point X with co-ordinates (x, y), where x=
,
y=
(3.2) .
3
(3.3)
We now catalogue the co-ordinates of the points O1, O2, O4 the circumcentres of triangles of triangles PQE, QRE, SPE and their corresponding orthocentres H1, H2, H4. The co-ordinates of O3, H3 for triangle RSE are omitted as results involving these points follow by similar reasoning. O1 (– ½mn(1 + t2), (1/4t)(m(1 + t2)){k(1 + t2) + n(1 – t2)}); O2 (– ½mn(1 + t2), (1/4t)(n(1 + t2)){m(1 – t2) – l(1 + t2)}); O4 ( ½kl(1 + t2), (1/4t)(k(1 + t2)){m(1 + t2) – l(1 – t2)}); H1 (km(1 – t2), – (1/2t)(m(1 – t2)){k(1 – t2) + n(1 + t2)}); H2 (– ln(1 – t2), (1/2t)(n(1 – t2)){l(1 – t2) – m(1 + t2)}); H4 (km(1 – t2), (1/2t)(k(1 – t2)){l(1 + t2) – m(1 – t2)}). 4. The remaining results Since Results 2 and 3 are known we do not provide further details, except to say that these results are a consequence of the fact that O1O2, O3O4 are perpendicular to QS, as are H1H4 and H3H2 all having infinite gradient, and the fact that the other four sides are perpendicular to PR having gradient
.
Result 4 We are now able to compute the squares of various lengths and find that O1O22 =
;
H1H22 =
;
O1O42 = H1H42 =
(4.1) (4.2)
;
(4.3)
;
(4.4)
From which we deduce O1O22/H1H22 = O1O42/H1H42 =
,
(4.5)
which, together with Results 2 and 3, establishes Result 4 that the two parallelograms are inversely similar. It is interesting to note that the enlargement factor of the similarity depends only on the angle between PR and QS. Result 5 Inspection of the co-ordinates given at the end of Section 3 shows that O1O2O3O4 is also in perspective with H3H4H1H2 with vertex of perspective the origin E. 4
Reference 1. C. J. Bradley, The Algebra of Geometry, Bath: Highperception (2007).
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Article 16 The Direct Similarity of the Miquel Point Configuration Christopher J Bradley 1.
Introduction
Given a triangle ABC and points L, M, N, other than the vertices, lying on BC, CA, AB respectively, then circles AMN, BNL, CLM share a common point P called the Miquel point. For a given triangle the position of P obviously depends on the points L, M, N. In this article we show that the centres X, Y, Z of the three circles form a triangle XYZ that is directly similar to ABC and that the centre of the similarity is the point P itself, by which we mean that XYZ is obtained from ABC by a rotation about P followed by dilation with centre P. We also give an analysis of the converse result. What this amounts to is that given a triangle ABC and a point P not on the sides, then it is always possible to find circles centres X, Y, Z to provide a configuration having P as a Miquel point, but the options are more limited than one might suppose. A direct similarity centre P is involved and any angle of rotation (other than a right angle) is possible, but then the scale factor of the dilation is fixed and depends on the angle of rotation. When the angle of rotation is 0° the scale factor is ½ and triangle LMN is the pedal triangle of P. We also consider the problem of where L, M, N have to be placed in order that the circles AMN, BNL, CLM should have equal radius. This problem results in equations that have to be solved numerically in order to find the relationship between the points L, M, N. Finally L, M, N are collinear if, and only if, P lies on the circumcircle of ABC and then AX, BY, CZ are concurrent, so that triangles XYZ and ABC are not only similar but in perspective. It follows from the work of Wood [1] on similar in perspective triangles that the perspector Q lies on both the circumcircles of ABC and XYZ. The circle XYZ may be identified as the circle of centres in the Wood configuration, see Bradley and Smith [2]. It therefore contains O, the circumcentre of ABC. In Fig. 5.1 we show the Miquel configuration for an arbitrarily chosen point P, when the angle of rotation of the similarity is 30°.
1
2. Figure arising from arbitrary L, M, N on the sides and the consequential similarity We start with a general, but economically parameterised triangle ABC, with A(0, 0), B(2, 2v), C(2, 2w), (w > v). The points L, M, N are now chosen on the sides BC, CA, AB respectively and to these are assigned co-ordinates L(2, 2u), M(h, hw), N(k, kv) (u ≠ v, w; h, k ≠ 0, 1). The equations of the circles BNL, CLM, AMN may be obtained using standard methods involving 4 x 4 determinants and are respectively x2 + y2 + (2(uv – 1) – k(v2 + 1))x – 2(u + v)y + 2k(v2 + 1) = 0, (2.1) 2 2 2 2 x + y + (2(uw – 1) – h(w + 1))x – 2(u + w)y + 2h(w + 1) = 0, (2.2) 2 2 2 2 2 2 (v – w)(x + y ) + (kw(v + 1) – hv(w + 1))x + (h(w + 1) – k(v + 1))y = 0. (2.3) It may be checked that these three circles share a common point P. We do not record its coordinates as they are complicated and are not needed in what follows. The co-ordinates of the centres of the three circles can now be determined and are 2
X : {1/(2(v – w))}( hv(w2 + 1) – kw(v2 + 1), k(v2 + 1) – h(w2 + 1)), Y : (½(k(v2 + 1) – 2(uv – 1)), u + v), Z : (½(hw2 + 1) – 2(uw – 1)), u + w). Calculations may now be carried out to show that XY/YZ = AB/BC = {√(v2 + 1)}/(w – v), and XZ/XY = AC/AB = √{(w2 + 1)/(v2 + 1)} and this establishes the similarity between triangle XYZ and ABC. We now consider the problem of when the three Miquel circles have a common area, which is when AL2 = BM2 = CN2. We find AL2 = (v2 + 1)(w2 + 1)(h2(w2 + 1) – 2hk(vw + 1) + k2(v2 + 1))/{4(v – w)2}, (2.4) 2 2 2 2 2 2 2 BM = ¼{k (v + 1) – 4k(v + 1)(uv + 1) + 4(u + 1)(v + 1)}, (2.5) 2 2 2 2 2 2 2 CN = ¼{h (w + 1) – 4h(w + 1)(uw + 1) + 4(u + 1)(w + 1)}. (2.6) Solving for h, k in terms of u we find the condition for equal areas is h = (2uv – v2 + 1)/(vw + 1), (2.7) and k = (2uw – w2 + 1)/(vw + 1). (2.8) l These formulas give the unique positions for M, N when L is given. A particularly obvious case is when h = k = 1 and u = ½(v + w), which is when P is the circumcentre and L, M, N the midpoints of the sides. No other case exists in which h = k. 3. Figure arising from choosing an arbitrary position for P and constructing a direct similarity with centre P In order to investigate the above direct similarity more fully we now investigate how a Miquel configuration can be constructed by choosing a point P to serve as a Miquel point and creating a direct similarity centre P for which the image of triangle ABC is a triangle XYZ so that X, Y, Z have the property that they are the centres of circles APMN, BPNL, CPLM where L, M, N lie on BC, CA, AC respectively. It is convenient, since rotation and dilation about P is to be effected, to choose P to be the origin, which means altering the co-ordinates of points. Now it is A, B, C and P that are fixed and the similarity that is specified. What happens is this. Triangle ABC is rotated about P by an angle θ to give a triangle A'B'C'. Now dilation is carried out by a scale factor d that has to be adjusted so that X lies on A'P and is also on the perpendicular bisector of AP. This is necessary for X to be the centre of a circle through A and P. Y and Z are then fixed by the direct similarity and they are then centres of circles through B and P and through C and P respectively. It turns out that although any angle θ (θ ≠ ½π) can be chosen, X, Y, Z do not lie on the required perpendicular 3
bisectors unless d = 1/(2cosθ). Rotation through a right angle is forbidden in order that X, Y, Z should be finite points. We choose the circumcircle of ABC to have radius 1 and equation (x – k)2 + y2 = 1.
(3.1)
Then the circumcentre has co-ordinates (k, 0) and P has co-ordinates (0, 0). Points on the circle can be assigned a real parameter ‘t’ so that they have co-ordinates {1/(1 + t2)}(t2(k – 1) + (k + 1), 2t) and to keep matters sufficiently general we choose A, B, C to have parameters a, b, c respectively. As indicated we now perform the direct similarity with rotation θ and scale factor d = 1/(2cosθ), so that the matrix that represents it has rows (½ , – s/(1 – s2)) and (s/(1 – s2), ½), where s = tan½θ. The resulting co-ordinates of X are {(1/(2(1 – s2)(1 + a2)}(a2(k – 1)(1 – s2) – 4as + (k + 1)(1 – s2), 2(a2s(k – 1) + a(1 – s2) + s(k + 1))). Those of Y, Z may be obtained by replacing a by b, c respectively. It may now be checked that XA = XP, YB = YP, ZC = ZP. Circles centres X, Y, Z through P now form a Miquel configuration in which pairs of circles meet at points (other than P) on the sides of the triangle. 4. What happens when P lies on the circumcircle of ABC What happens when P lies on the circumcircle is that the previous analysis may be used with k = 1. A now has co-ordinates {2/(a2 + 1)}(1, a) with similar expressions for B and C. The circumcircle ABC has equation x2 + y2 – 2x = 0. (4.1) X has co-ordinates {1/((a2 + 1)(s2 – 1))}(2as + s2 – 1, a(s2 – 1) – 2s) with similar expressions for Y and Z. The circle centre X passing through A has equation (a2 + 1)(s2 – 1)(x2 + y2) – 2(2as + s2 – 1)x – 2(a(s2 – 1) – 2s)y = 0. (4.2) The circle centre Y passing through B has a similar equation with b replacing a. These two circles meet at P and at the point N with co-ordinates {1/((s2 – 1)(a2 + 1)(b2 + 1))}(2(ab – 1)(1 – s2) + 2s(a + b), 2(2s(ab – 1) + (a + b)(s2 – 1)), with similar expressions for L and M. It is straightforward to show these points lie on the sides of ABC. Forming the 3x 3 determinant whose rows are the co-ordinates of L, M, N (with third element 1) we find that it vanishes, showing that LMN is a straight line. The equation of AX is 4
(a(s2 – 1) + 2s)x + (2as – (s2 – 1))y – 4s = 0.
(4.3)
This line meets the circumcircle with Equation (3.1) at the point Q with co-ordinates {(4s/(s2 + 1)2}(2s, 1 – s2). As this is independent of a, lines BY, CZ also pass through Q, which means that triangles ABC and XYZ are also in perspective. Circle XYZ has equation (s2 – 1)(x2 + y2) + (1 – s2)x + 2sy = 0.
(4.4)
This circle passes through P and also through O, the centre of the circumcircle of ABC, showing that circle XYZ is the circle of centres in the Wood [1] configuration based on ABCPQ. It may also be shown that LMN is the double Wallace-Simson line of P with respect to triangle XYZ triangle. Since the construction is reversible this demonstrates that every transversal of ABC is the Double Simson line of a point P on its circumcircle with respect to some triangle XYZ that is directly similar to ABC with centre of similarity P. See Fig. 2 for a record of these results.
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Fig. 2
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Figure 3 5. Paralogic Wood configuration This is a very natural construction. Start with a triangle ABC and a transversal LMN, with L on BC etc. Now erect perpendiculars at L, M, N to BC, CA, AB respectively to create by their intersections in pairs the triangle XYZ. The ABC and XYZ are directly similar and in perspective with LMN the Desargues’ axis and the circles ABC and XYZ meet at the perspector K and the centre of inverse symmetry J. Moreover the circles are orthogonal, so the triangles ABC and XYZ are paralogic. The paralogic centres are the orthocentres of the two triangles. See Figure 3.
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Reference 1. F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 17 Harmonic ranges in a coaxal system of circles Christopher J Bradley
1. Introduction Suppose we are given a circle that cuts the sides BC, CA, AB of a triangle ABC at pairs of points on each side, say P and L on BC, Q and M on CA and R and N on AB, in each case in the specified order working around the triangle in an anticlockwise direction. Now let P' and L' be the harmonic conjugates of P and L with respect to B and C, Q' and M' the harmonic conjugates of Q and M with respect to C and A and R' and N' the harmonic conjugates of R and N with respect to A and B. First we prove that P', L', Q', M', R' and N' always lie on a conic. We then investigate the conditions on the given circle for this conic also to be a circle. We define such a pair of circles as a harmonic pair of circles and it turns out that each such pair belongs to the coaxal system of circles of which the polar circle, centre the orthocentre H of ABC and the circumcircle, centre the circumcentre O of ABC may be regarded as defining members. Since the polar circle is real only when ABC is obtuse we remark that other circles in this coaxal system are the nine-point circle, centre T, the midpoint of OH, and the orthocentroidal circle based on GH as diameter, where G is the centroid of ABC. If a harmonic pair S and S' have centres at X and X', then X and X' divide O and H harmonically. See Fig. 1. Investigation of the equations of the circles forming a harmonic pair enables us to relate their equations in a canonical form in terms of those of the polar circle and the circumcircle, with the position of their centres. Analysis that is first applied to the coaxal system of circles with centres on the Euler line may be transferred to the circles of any coaxal system. If one sets up a more general figure in which conics replace circles and one draws the lines PN, QL, RM to form a triangle DEF and one draws the lines P'N', Q'L', R'M' to form a triangle UVW then we prove that triangles ABC, DEF, UVW are mutually in perspective, and moreover the three Desargues’s lines of the three perspectives coincide and the three perspectors are collinear. This is possibly a known result, but as its proof is straightforward we include it thereby providing an extension of the results about harmonic pairs of circles. 1
Areal co-ordinates are used throughout the article.
2. -Harmonic conjugate points lie on a conic Suppose a conic cuts the sides BC, CA, AB of a triangle in points P, L on BC, Q, M on CA and R, N on AB, then by Carnot’s theorem (BP/PC)(BL/LC)(CQ/QA)(CM/MA)(AR/RB)(AN/NB) = 1. (2.1) If P' is the harmonic conjugate of P with respect to B and C, and L', Q', M', R', N' are similarly defined, then we have (BP'/P'C) = – (BP/PC) etc. And so from Equation 1.1, (BP'/P'C)(BL'/L'C)(CQ'/Q'A)(CM'/M'A)(AR'/R'B)(AN'/N'B) = 1. (2.2)
2
Now, if L, M, N and P, Q, R are the feet of two Cevians, then L'M'N' and P'Q'R' are two straight lines, by the converse of Menelaus’s theorem, and hence form a degenerate conic. Otherwise the conditions for the converse of Carnot’s theorem are satisfied and P', L', Q', M', R', N' lie on a non-degenerate conic. The degenerate case occurs in at least one familiar case when the conic is the nine-point circle, and the harmonic conjugates then lie on two straight lines, one being the line at infinity. This is perhaps a good place to mention that if we work in the complex field the theorems in this article are true irrespective of whether the points of intersection of the conic cut the triangle sides in real points or not. 3. The condition for the harmonic conjugate points to lie on a circle Suppose that P and L have co-ordinates (0, q, r) and (0, m, n) respectively. Then if we put x = 0 in the equation of the given conic it must reduce to (ry – qz)(ny – mz) = 0. Since P' and L' are the harmonic conjugates of P and L, their co-ordinates must be (0, q, – r) and (0, m, – n) and it follows that if we put x = 0 in the equation of the harmonically related conic it must reduce to (ry + qz)(ny + mz) = 0. Thus under harmonic conjugation rny2 – (rm + qn)yz + qmz2 becomes rny2 + (rm + qn)yz + qmz2 and the change is effected simply by altering the sign of the yz term. If then the original conic has equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
it follows immediately that the harmonically related conic has equation ux2 + vy2 + wz2 – 2fyz – 2gzx – 2hxy = 0.
(3.2)
We now investigate the conditions under which both these equations represent circles. From Section 2.3.10 of Bradley [1] there must exist a constant k such that the following six equations hold. v + w – 2f = a2, w + u – 2g = b2, u + v – 2h = c2 v + w + 2f = –ka2, w + u + 2g = –kb2, u + v + 2h = –kc2. (3.3) The solutions of Equation (2.3) are f = – ¼a2(k + 1), g = – ¼b2(k + 1), h = – ¼c2(k + 1), u = ¼(b2 + c2 – a2)(– k + 1), v = ¼(c2 + a2 – b2)(– k + 1), w = ¼(a2 + b2 – c2)(– k + 1).
(3.4)
From Section 2.3.6 of Bradley [1] the co-ordinates of the centre of the conic with Equation (3.1) are proportional to (vw – gv – hw –f2 + fg + fh, wu – hw – fu – g2 + gh + gf, uv – fu – gv –h2 + hf + hg).
3
Substituting values from Equation (3.4) we find the x-co-ordinate to be proportional to –2a4k + a2(1 + k)(b2 + c2) – (–k + 1)(b2 – c2)2 with y- and z-co-ordinates found from (3.5) by cyclic change of a, b, c. Now the equation of the Euler line of triangle ABC is (b2 – c2)(b2 + c2 – a2)x + (c2 – a2)(c2 + a2 – b2)y + (a2 – b2)(a2 + b2 – c2)z = 0.
(3.5)
(3.6)
Substituting the co-ordinates of the centre of the circle into Equation (2.6) we find the centre lies on this line. Similar analysis holds for the conic with Equation (3.2). It follows that whatever the value of k the centres of the two circles must lie on the Euler line. 4. The coaxal system formed by harmonic pairs of circles From Section 3 we know that the equations of a harmonic pair of circles are of the form (– k + 1){(b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2} – 2(k + 1){a2yz + b2zx + c2xy} = 0, and (– k + 1){(b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2} + 2(k + 1){a2yz + b2zx + c2xy} = 0, Note that one may obtain Equation (4.2) from Equation (4.1) by replacing k by 1/k.
(4.1)
(4.2)
First let us look at some particular cases. When k = – 1 one gets the polar circle with equation (b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2 = 0 (4.3) from both Equations (4.3) and (4.4). It follows that the polar circle is self-conjugate being its own harmonic partner. Note that the polar circle is real if and only if triangle ABC is obtuse, so if one wishes to visualise cases for all real k one must draw ABC as obtuse. Similarly if one puts k = 1 one gets the circumcircle with equation 2(a2yz + b2zx + c2xy) = 0, (4.4) again from both Equations (4.1) and (4.2). (The factor 2 is introduced in Equation (4.4) rather than in Equations (4.7) and (4.8) below.) It follows that the circumcircle is also self-conjugate. If one puts k = 0, Equation (4.1) becomes the nine-point circle with equation {(b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2}– 2{a2yz + b2zx + c2xy} = 0. (4.5) Equation (4.2) factorizes and reduces to the line pair x + y + z = 0 and (b2 + c2 – a2)x + (c2 + a2 – b2)y + (a2 + b2 – c2)z = 0. The first of these is the line at infinity and the second line we shall identify very soon.
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(4.6)
Abbreviating equations (4.3) and (4.4) to read SP = 0 and SC = 0 we see that Equations (4.1) and (4.2) may be written in the form (1 – k)SP – (1 + k)SC = 0 (4.7) and (1 – 1/k)SP – (1 + 1/k)SC = 0 (4.8) respectively. It follows that a pair of circles are harmonic if, and only if, their centres are on the Euler line and they are in the coaxal system of circles defined by the polar circle and the circumcircle. For an obtuse-angled triangle ABC, these two circles define an intersecting system of coaxal circles and the second of Equations (4.6) is therefore their common chord. In the case of an acute-angled triangle, when working over the real field, one may view the coaxal system as a non-intersecting system of circles, as in Fig.1, defined by the circumcircle and the line corresponding to the second of Equations (4.6), which is now the radical axis.
5. The location of the centres of a harmonic range of circles Circle centres may be computed using the formula from Section 2.3.6 of Bradley [1]. Details of the calculation, carried out by DERIVE, are omitted. Only the x-co-ordinate of each point needs to be recorded, since the y- and z-co-ordinates follow by cyclic change of a, b, c. Co-ordinates are normalized, as is necessary for deductions to be made about the disposition of the centres. For the polar circle the centre is at H with x-co-ordinate xP = (a2 + b2 – c2)(a2 + c2 – b2)/(a + b + c)(b + c – a)(c + a – b)(a + b – c).
(5.1)
For the circumcircle the centre is at O with x-co-ordinate xC = a2(b2 + c2 – a2)/(a + b + c)(b + c – a)(c + a – b)(a + b – c).
(5.2)
For the circle with Equation (4.1) the x-co-ordinate is xk = (1/2)(1 – k)xP + (1/2)(1 + k)xC.
(19)
For the circle with Equation (4.2) the x-co-ordinate is x(1/k) = (1/2)(1 – 1/k)xP + (1/2)(1 + 1/k)xC.
(20)
Since {xP, xC; xk, x(1/k)} = {– 1, 1; k, 1/k} = – 1 the centres of any harmonic pair of circles separate the centre of the polar circle and the centre of the circumcircle harmonically. Another circle in this coaxal system is the orthocentroidal circle, which is the circle with GH as diameter. It has parameter k = – 1/3. Circles centre G, the centroid, and deL, deLongchamps point, form a harmonic pair with k = 1/3 and k = 3 respectively. 5
6. Perspectives formed by harmonic pairs of circles The results of this section do not depend on the conics involved being circles, so we establish the results in the more general case. The notation used, however, is the same as before, with a conic meeting sides BC, CA, AB respectively at pairs of points P, L; Q, M, R, N. The harmonic conjugate points are denoted by P', L', Q', M', R', N' and as we have seen in Section 2 these points also lie on a conic. Suppose now that lines PN, QL, RM are drawn to form triangle DEF with D = PN^QL etc. and lines P'N', Q'L', R'M' are drawn to form triangle UVW with U = P'N'^Q'L' etc. Result 1
Triangles ABC and DEF are in perspective
Proof BC^EF is the same as RM^PL = I, say. CA^FD is the same as QM^PN = J, say. AB^DE is the same as RN^QL = K, say. It follows, by Pascal’s theorem for the hexagon (RMQLPN) inscribed in the conic S that IJK is a straight line, which is therefore the Desargues’s axis of perspective for triangles ABC and DEF. Using the notation of Fig. 2 the vertex of perspective is the point Z. Result 2
Triangles ABC and UVW are in perspective
Proof Since P'L' is the same line as PL, Q'M' is the same line as QM and R'N' is the same line as RN the proof of Result 2 follows by the same reasoning as for Result 1, but with the hexagon (R'M'Q'L'P'N') inscribed in the conic S'. Furthermore the intersections are again the points I, J, K which is the same Desargues’s axis as before. The vertex of perspective is shown as the point Y in Fig. 2. Result 3
Triangles DEF and UVW are in perspective
Proof Since VW^EF = RM^R'M' = I etc., Result 3 is immediate and once again the Desargues’ axis is the same line. The vertex of perspective is shown as X in Fig. 2. Result 4
The perspectors X, Y, Z are collinear
Proof By using a projective transformation we may map the line IJK to the line at infinity and then the three triangles map into three triangles homothetic in pairs. The question now reduces to whether the three homothety centres are collinear. Denote the three pairwise homothetic triangles by T1, 6
T2, T3 and let the homothety θij carrying Ti to Tj have centre Cij. Now θ12 followed by θ23 leaves the line joining C12C23 invariant. This composition is θ13 so C13 is on this line.
I am grateful to Dr Geoff. Smith of the University of Bath for providing the proof of Result 4 and for having a number of useful conversations with me concerning the contents of this article. Reference 1. C.J.Bradley, The Algebra of Geometry, Highperception, Bath 2007.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 18 Some Special Circles in a Triangle Christopher J Bradley
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Introduction 1
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We study a construction in a triangle ABC that assigns to any point P, not on the sides, circumcircle or altitudes of a triangle, a unique circle passing through P. This is done by finding three points U, V, W that define the circle and once it has been shown that U, V, W, P are concyclic, then four other points Q, X, Y, Z are determined with the property that UX, VY, WZ are concurrent at Q. The construction is as follows. Choose P and draw the circles BPC, CPA, APB, which we denote by Γ1, Γ2, Γ3 respectively. Now draw AP, BP, CP to meet Γ1, Γ2, Γ3 respectively at D, E, F. Draw through D the perpendicular to BC to meet Γ1 again at U, with V, W defined similarly on perpendiculars to CA, AB respectively. Then it turns out that U, V, W, P are concyclic and lie on a circle we denote by ΓP. Now draw AP, BP, CP to meet ΓP at X, Y, Z respectively. Then it may be proved that UX, VY, WZ are concurrent at a point Q. Furthermore AQ, BQ, CQ intersect BC, CA, AB respectively at points L, M, N that also lie respectively on DU, EV, FW. Those familiar with the construction of Hagge circles, see Bradley and Smith [1], will recognize the similarity that exists between the two constructions, except that Hagge circles define all circles passing through the orthocentre H, whereas our construction gives one circle (for any given triangle) passing through all points P not lying on the sides or altitudes of ABC. See the figure above for an illustration of the construction. In the analysis that follows we take P to be the centroid G, so all that is proved is that the construction works in this case. However, the computer geometry software package CABRI II plus is so accurate that we may be sure that the construction holds generally. However, the algebra involved for a general point is so formidable (even with the aid of a computer algebra software package) that we offer this as a most potent excuse for not covering the general case. We use areal co-ordinates throughout, an account of which is given by Bradley [2, 3]. 2.
The circles BGC, CGA, AGB and the points D, E, F
The equation of a circle using areal co-ordinates is always of the form (2.1) where u, v, w are constants to be determined and a, b, c are the side lengths of ABC. To find the equation of circle BGC we put the co-ordinates of points B, G, C in Equation (2.1) and get three equations to determine u, v, w. The values obtained are
and the
equation of circle BGC is accordingly (2.2)
2
The equations of circles CGA, AGB may be obtained by cyclic change of x, y, z and a, b, c. The equation of AG is y = z and this meets the circle BGC at the point D with co-ordinates D(– 3a2, (a2 + b2 + c2), (a2 + b2 + c2)). Points E, F have co-ordinates that may be obtained from those of D by cyclic change of x, y, z and a, b, c. 3.
Lines through D, E, F perpendicular to BC, CA, AB
Finding perpendicular lines when using areal co-ordinates is tiresome, but the results are known, see [3], and may therefore be quoted. If we take a point T with co-ordinates (d, e, f) then the foot of the perpendicular from T on the line BC has co-ordinates
and consequently the equation
of the line perpendicular to BC, x = 0, through the point T has equation (3.1) If we now put d = –3a2, e = f = (a2 + b2 + c2), we get the equation of DU, which is (3.2) The equations of EV and FW may be written down from Equation (3.2) by cyclic change of x, y, z and a, b, c. 4.
Points U, V, W and the circle through G
The line DU, with Equation (3.2), meets Γ1 with Equation (2.2) at the point U with co-ordinates (x, y, z), where (4.1)
The co-ordinates of the points V, W may be found by cyclic change of x, y, z and a, b, c. We now substitute the co-ordinates of U, V, W into Equation (2.1) to obtain three equations for u, v, w. These values are then substituted back in Equation (2.1) to obtain the unpromising equation of the circle ΓG, which is
.
(4.2) 3
Terms in y2, z2, and zx, xy may be obtained by cyclic change of a, b, c of those in x2 and yz. It may now be checked that this circle passes through G(1, 1, 1). 5.
The points X, Y, Z and the linking point Q
The equation of AG is y = z and this meets the circle ΓG with Equation (4.2) at the point X with co-ordinates (x, y, z), where (5.1)
The co-ordinates of Y, Z, where BG, CG respectively meet Γ G may be obtained from those of X by cyclic change of x, y, z and a, b, c. The equations of the lines UX, VY, WZ are lengthy and no good purpose would be served by writing them down. However, the algebra computer package DERIVE, which we used throughout, showed that the three lines meet at the point Q, whose co-ordinates (x, y, z) are given by (5.2) The apparent simplicity of this result is very gratifying. We use the term linking point for Q, as it links the point G with the points lying on circle ΓG. This it does in two ways. The first has just been explained as Q is the point of concurrency of UX, VY, WZ. The second way is that the lines AQ, BQ, CQ meet the lines DU, EV, FW respectively at points L, M, N lying on BC, CA, AB respectively. To see this, note that the equation of AQ is (5.3) and from Equation (3.2) it can be seen that DU meets BC at this point also. Note that Q may lie outside triangle ABC. References 1. C. J Bradley & G.C.Smith, Math. Gaz. pp202 -207 July 2007. 2. C. J. Bradley, Challenges in Geometry, Oxford (2005). 3. C. J .Bradley, The Algebra of Geometry, Highperception, Bath (2007). Flat 4, Terrill Court, 12/14 Apsley Road, BRISTOL BS8 2SP 4
Article 19 Circular Perspective Christopher J Bradley 1. Introduction If two triangles ABC and PQR are such that AP, BQ, CR are concurrent at a point V then the triangles are said to be in perspective. The vertex V is sometimes called the perspector. This relation is obviously a symmetric property. When the triangles are in perspective then the dual property also holds that L = BC^QR, M = CA^RP, and N = AB^PQ are collinear. The line LMN is called Desargues’ axis of perspective and more recently is sometimes called the perspectrix. The main theorem is that if triangles are in double perspective so that ABC is in perspective, say, with both triangles PQR and QRP, then they are bound to be in triple perspective, meaning that triangle ABC is in perspective with triangle RPQ as well. There is also the concept of reverse perspective when one considers ABC in relation to triangles PRQ, RQP or QPR. For example, pairs of triangles in the Brocard porism are in perspective with perspector the symmedian point K and are in triple reverse perspective with the three perspectors lying on the polar line of K. In this paper we introduce a concept which we call circular perspective. This is a relation between a pair of triangles ABC and PQR in which the vertices of both triangles possess the property that vertices of one do not lie on the sides or extended sides of the other. The definition of circular perspective is as follows: ABC and PQR are in circular perspective if circles BCP, CAQ and ABR have a common point D. It is by no means obvious that this is a symmetric relation, but we prove that this is the case. In fact we prove two main theorems: (i) If ABC is in circular perspective with PQR, then PQR is in circular perspective with ABC. (ii) If ABC is in circular perspective with both PQR and QRP then it is in circular perspective with triangle RPQ. We conclude our study by investigating an interesting case in the geometry of the triangle, involving the orthocentroidal circle and the Brocard points, in which triple circular perspective exists. Note that there is no concept of reverse perspective since, for example, circle CPQ is the same as circle CQP. The theorem stated in (i) is illustrated in Figure 1. 2. Circular perspective is a symmetrical relation The proof of (i) is straightforward. If D is the point of concurrence of circles BCP, CAQ and ABR, then invert with respect to D and the configuration becomes one in which (using stars for inverted points) P* lies on B*C*, Q* lies on C*A* and R* lies on A*B*. It then follows that
1
circles Q*R*A*, R*P*B*, P*Q*C* are concurrent at the point S*, the Miquel point. The inverse image S of the point S* is now the point of concurrence of circles QRA, RPB and PQC.
A
Q P D R B C S
Figure 1
3. Double circular perspective implies triple circular perspective Suppose now that the circles BCP, CAQ, ABR meet at D and circles BCQ, CAR, ABP meet at E. We want to prove that circles BCR, CAP, ABQ meet at a point F. As in Section 2 we may invert with respect to D, and dropping the stars on invoice points it is sufficient to prove that if P, 2
N
Q, R lie on BC, CA, AB respectively and circles BCQ, CAR, ABP meet at a point E, then circles BCR, CAP, ABQ meet at a point F. This result is illustrated in Figure 2. We prove this result using areal co-ordinates with ABC as triangle of reference and supposing P has co-ordinates (0, p, 1 – p), Q has co-ordinates (1 – q, 0, q) and R has co-ordinates (r, 1 – r, 0).
Q
A
E
R F
B
C
P
Figure 2
The equation of any circle using areal co-ordinates is a2yz + b2zx + c2xy – (x + y + z)(ux + vy + wz) = 0.
(3.1)
To obtain the equation of circle ABP we insert into Equation (3.1) the co-ordinates of A, B, P in succession and obtain three equations for u, v, w. The solution is then substituted back into Equation (3.1). The result is u = v = 0 and w = a2p and consequently the circle ABP has equation 3
– a2pz2 + a2(1 – p)yz + (b2 – a2p)zx + c2xy = 0.
(3.2)
The equations of the circles BCQ, CAR may now be written down by cyclic change of x, y, z and a, b, c. Circles ABP and BCQ meet at the point E with co-ordinates (x, y, z), where x = a2p(c2p + b2q(1 – p)), y = b2pq(a2p – b2(1 – q)), z = b2q(c2p + b2q(1 – p)).
(3.3)
If we now put in the condition that circle CAR also passes through E we get an expression (equal to zero) that factorizes and discarding one factor that cannot hold because p and q are real, we find the condition for concurrence to be r(1 – q)/a2 + p(1 – r)/b2 + q(1 – p)/c2 = 0. (3.4) We now consider the circles ABQ, BCR, CAP. The equation of circle ABQ is – b2(1 – q)z2 + (a2 + b2(1 – q))yz + b2qzx + c2xy = 0.
(3.5)
The equations of circles BCR and CAP may now be written down by cyclic change of x, y, z and a, b, c. Circles ABQ and BCR meet at the point F with co-ordinates (x, y, z), where x = b2(1 – q)(a2(1 – r) + b2r(1 – q)), y = b2(1 – q)(1 – r)(c2(1 – r) – b2q)), z = c2(1 – r)(a2(1 – r) + b2r(1 – q)).
(3.6)
The condition that F also lies on circle CAP, after similar algebra as before, is again found to be Equation (3.4), which completes the proof. When properties (i) and (ii) are taken into account it follows that if ABC and PQR are in double circular perspective then both ABC and PQR, and PQR and ABC are in triple circular perspective. It is also the case in projective geometry that if ABC and PQR are two triangles and X, Y are two distinct points then if conics BCPXY, CAQXY, ABRXY have a common point Z then we may say that triangles ABC and PQR are in conical perspective by means of triangle XYZ. Generalizations of properties (i) and (ii) now hold by relating X and Y to the circular points at infinity. 4. An example of triple circular perspective
4
Consider the orthocentroidal circle on GH as diameter, where H is the orthocentre of ABC and G is its centroid. Its equation, see Bradley and Smith [1], is (b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2 – a2yz – b2zx – c2xy = 0.
(4.1)
The median AG, with equation y = z, meets this circle at a point we call aH and its co-ordinates are easily calculated to be aH(a2, b2 + c2 – a2, b2 + c2 – a2). The points bH and cH are similarly
A
M
bH G
' cH aH H L B
C
N
Figure 3
defined and have co-ordinates bH(c2 +a2 – b2, b2, c2 + a2 – b2) and cH(a2 + b2 – c2, a2 + b2 – c2, c2). What we show is the following: 5
Triangles ABC and aHbHcH are in triple circular perspective. The method of proof is as follows: (i) We show that circles ABaH, BCbH, CAcH all pass through the Brocard point Ω with co-ordinates (1/b2, 1/c2, 1/a2). (ii) We then show that circles ABbH, BCcH, CAaH all pass through the Brocard point Ω' with co-ordinates (1/c2, 1/a2, 1/b2). (iii) It then follows by property (ii) that the circles ABcH, BCaH, CAbH all pass through a fixed point, which we show is the orthocentre H. (iv) It then follows by property (i) that triangles ABC and aHbHcH are in triple circular perspective. (v) As a corollary it then follows that circles BaHcH, CbHaH, AcHbH have a common point L, circles BaHbH, CbHcH, AcHaH have a common point M and circles BcHbH, CaHcH, AbHaH have a common point N. These results are illustrated in Figure 3. Proof of (i) The equation of the circle BCbH is b2x2 + (b2 – c2)xy – a2yz = 0.
(4.2)
It is easily checked that this circle passes through Ω(1/b2, 1/c2, 1/a2). The equations of circles CAcH, ABaH may be written down from Equation (4.2) by cyclic change of x, y, z and a, b, c. And it may now be checked that both these circles also pass through Ω. Proof of (ii) The equation of the circle BCcH is c2x2 – a2yz + (c2 – b2)zx = 0.
(4.3)
It is easily checked that this circle passes through Ω'(1/c2, 1/a2, 1/b2). The equations of circles CAaH, ABbH may be written down from Equation (4.3) by cyclic change of x, y, z and a, b, c. And it may now be checked that both these circles also pass through Ω'. Proof of (iii) It now follows from Section 3 that the circles ABcH, BCaH, CAbH all pass through a fixed point, but it remains to show that fixed point is H. In fact the equation of circle BCaH is a2yz + b2zx + c2xy – (b2 + c2 – a2)x(x + y + z) = 0. (4.4) 2 2 2 2 2 2 It may now be checked that this circle passes through H(1/(b + c – a ), 1/(c + a – b ), 1/(a2 + b2 – c2)). 6
Proof of (iv) It now follows by Section 2 that triangles ABC and aHbHcH are in triple circular perspective. Proof of (v) As a corollary it follows immediately from (iv) that circles BaHcH, CbHaH, AcHbH have a common point L, circles BaHbH, CbHcH, AcHaH have a common point M and circles BcHbH, CaHcH, AbHaH have a common point N. We do not find the co-ordinates of L, M, N, as it is really only their existence which is significant. In any case experience suggests that the algebra computer package DERIVE, which we use, would not be able to find them, the solving of algebraic (rather than numerical) simultaneous quadratics not being possible unless they are very straightforward as in Section 3. It may just be added that there is a great deal more to this configuration, once the positions of the centres of the eighteen circles have been identified. Many similarities exist, but that is another story. Dedication This article is dedicated to my great nephew Matthew Joshua Bradley, it having been composed during the first two weeks of his life. Reference 1. C.J.Bradley and G.C.Smith, The locations of triangle centres, Forum. Geom., 6 (2006) 57-70. Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
7
Article 20 On the Nine Intersections of Two Cevian Triangles Christopher J Bradley
1. Introduction Let DEF and LMN be the Cevian triangles of distinct points S and T, with D, L on BC, E, M on CA and F, N on AB. It is presumed that S and T do not lie on a sideline of triangle ABC nor on a line parallel to a sideline through an opposite vertex. We introduce the nine intersections of these two triangles: P = EF^MN, Q = FD^NL, R = DE^LM; X = DE^NL, Y = EF^LM, Z = FD^MN; U = FD^LM, V = DE^MN, W = EF^NL. 1
See the figure above for an illustration in which S and T are internal points, but the following results are true whether or not S and T are internal. We have four triangles ABC, PQR, XYZ, UVW and together with the two Cevian triangles they exhibit a number of properties, some of which are quite striking. These properties, which we establish in the following sections using areal (barycentric) coordinates - see [1], [2], are as follows: (i) A, Q, R are collinear as are B, R, P and C, P, Q; (ii) BC, EN, YZ are concurrent at a point A', CA, FL, ZX are concurrent at a point B' and AB, DM, XY are concurrent at a point C'; (iii) BC, MF, VW are concurrent at a point A'', CA, ND, WU are concurrent at a point B'', and AB, LE, UV are concurrent at a point C''; (iv) A', B', C' are collinear, A'', B'', C'' are collinear and these two lines are concurrent with ST at a point O; (v) A, X, U are collinear as are B, Y, V, and C, Z, W. (vi) (v) implies that triangles ABC, XYZ, UVW are mutually in perspective having a common perspector J; (vii) In fact each of these triangles is also in perspective with triangle PQR. Triangles ABC and PQR with perspector H (off the page of the figure), triangles PQR and XYZ with perspector I and triangles PQR and UVW with perspector K; (viii) The perspectors I, J, K are collinear. 2. Setting the scene Let the Cevian points S and T have co-ordinates (d, e, f) and (l, m, n) respectively. The equations of their sides are straightforward to obtain and the co-ordinates of the nine points of intersection introduced in Section 1 are as follows:
) ) . 3. Result (i): Triangle ABC is inscribed in triangle PQR
2
Since the y- co-ordinates of Q and R are identical, as are also their z- co-ordinates it follows that the line QR passes through A. Similarly the line RP passes through B and the line PQ passes through C. 4. Results (ii) and (iii): Pairs of lines concurrent at points on the sides of triangle ABC The equation of the line YZ is .
(4.1)
The equation of the line EN is (4.2) These meet at the point A' with co-ordinates (0, – dm, fl) and A' lies on BC. Similarly ZX and FL meet at a point B'(dm, 0, – en) on CA, and XY and DM meet at a point C'(– fl, en, 0) on AB. Likewise the equation of VW is (4.3) The equation of MF is (4.4) These meet at the point A'' with co-ordinates (0, – el, dn) and A'' lies on BC. Similarly WU and ND meet at a point B''(el, 0, – fm) on CA, and UV and LE meet at a point C''(– dn, fm, 0) on AB. 5. Result (iv): Three concurrent lines Points A', B', C' are collinear on the line with equation (5.1) Points A'', B'', C'' are collinear on the line with equation (5.2) These meet at the point O with co-ordinates
The equation of ST, the line joining the Cevian points, is (5.3) It may now be checked that O lies on the line ST.
3
6. Results (v) and (vi): Three perspectives with the same perspector The lines AUX, BVY, CWZ have equations (6.1) (6.2) (6.3) respectively. These lines are concurrent at the point J with co-ordinates
It follows that triangles ABC, UVW, XYZ are mutually in perspective with J as the common perspector. 7. Result (vii): Triangle PQR is in perspective with all of ABC, UVW, XYZ First consider the pair of triangles ABC and PQR. The equations of AP, BQ, CR are (7.1) (7.2) (7.3) respectively. These lines are concurrent at the point H with co-ordinates
Thus triangles ABC and PQR are in perspective with H as perspector. Note that as triangle ABC is inscribed in triangle PQR (Result (i)), it follows that H is a Cevian point of triangle PQR with ABC as its Cevian triangle. This point has recently featured in a paper by Pohoata and Yiu [3], who also show that the corresponding sidelines of three Cevian triangles are concurrent if and only if the three Cevian points are coconic with the vertices of triangle ABC. They call H the anticevian point of S and T. Next consider the pair of triangles PQR and XYZ. The equations of PX, QY, RZ are (7.4) (7.5) (7.6) respectively. These meet at the point I with co-ordinates (x, y, z), where
4
Thus triangles PQR and XYZ are in perspective with perspector I. Next consider the pair of triangles PQR and UVW. The equations of PU, QV, RW are (7.7) (7.8) (7.9) respectively. These meet at the point K whose co-ordinates are (x, y, z), where ,
Thus triangles PQR and UVW are in perspective with perspector K. 8. Result (viii): The perspectors I, J, K are collinear It may now be checked that the three perspectors I, J, K lie on the line with equation .
(8.1)
9. An alternative construction for the anticevian point H As is well known the harmonic conjugates of the feet of the Cevians of a point are collinear on a transversal associated with the point. The equation of the transversal associated with S(d, e, f) is x/d + y/e + z/f = 0, (21) and it is easily checked that this line contains the point H In fact H lies on the transversal associated with any point on the circumconic ABCST, and as shown in [3] the triangle PQR is also invariant for all pairs of Cevian triangles involving pairs of points lying on this circumconic. References 1. C. J. Bradley, The Algebra of Geometry, Bath: Highperception (2007). 2. C. J. Bradley, Challenges in Geometry, Oxford: University Press (2005). 3. C. Pohoata & P.Yiu, On a product of two points induced by their Cevian triangles, Forum Geometricorum, 7 (2007) 169-180. Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP 5
Article 21 Significant Points on Circles Centre the Circumcentre Christopher J Bradley
A
P
X
N
M
V
O
E
F
W Z
Q Y
C
L
B U
D
Figure 1
1.
Introduction
Given a triangle ABC with circumcentre O and a point P not on its sides or their extensions and not on the circumcircle, it is shown that one may construct on the circle centre O and radius OP six significant points. The construction of U, V, W are straightforward enough; they are the 1
second intersections of AP, BP, CP with the circle. The other three points X, Y, Z arise in a somewhat elaborate fashion. The point X, for example, arises as follows: Join P to the midpoint L of BC and extend it to D, where L is the midpoint of PD. Then reflect D in the side BC to produce X. Points Y, Z follow similarly using the midpoints M, N of the sides CA, AB respectively. Alternatively, and more easily, X is the reflection of P in the perpendicular bisector of BC. Two very interesting properties arise. First, the lines UX, VY, WZ are concurrent at a point Q. And secondly, triangle XYZ is inversely similar to triangle ABC. We prove the second result by establishing two lines of inverse similarity passing through O, with the property that ABC may be mapped into XYZ by using one of these lines and O as a centre of inverse spiral symmetry. The first property means that the inverse image T of Q under this transformation must be such that AD', BE', CF' are concurrent at T, where D', E', F' are the inverse images of U, V, W respectively. As is well known, the point T may now be used to manufacture a circle passing through the orthocentre H, known as a Hagge circle, see [1, 2]. The Hagge circle is also obtained by an inverse similarity of triangle ABC with T the centre of inverse similarity, and it follows from our analysis that the circle through P and the points on it are related to those on the Hagge circle by a direct congruence, which is just a rotation of 180o about a point midway between the centres. We do not repeat any of the work on the derivation of Hagge circles in this paper, but it is our purpose to point out the connections that exist. The procedure may, of course, be reversed. Start with a Hagge circle and rotate by 180o in a way that will be explained and you end up with a circle centre O passing through the isogonal conjugate of the point that is used to construct the Hagge circle. The article concludes with a conjecture (verified by CABRI, but without algebraic or geometrical proof) that other direct congruences not only map the Hagge circle and its points on to other circles with similar properties (which must obviously be the case), but the new circles manufactured have additional properties that make them interesting and significant. We believe this removes the mystique of why circles through the orthocentre are somehow thought to be special. They are no more special than any other circle once the appropriate conjugation and direct congruence are identified. In this article we produce an analysis to prove the assertions made about our construction, using Cartesian co-ordinates. Care is needed in following the text to refer to the appropriate figure, as changes of notation from one section to the next are inevitable, so many points being involved. 2. The six points X, Y, Z, U, V, W Let the circumcircle have equation x2 + y2 = 1, with centre O(0, 0) and suppose P has coordinates (p, 0). For the co-ordinates of A we use the familiar expressions (2a/(1 + a2), (1 – a2)/(1
2
+ a2)), and similar for B and C with parameters b and c respectively. In this section and the next the notation is that of Figures 1 and 2. L, the midpoint of BC, has co-ordinates (k(b + c), k(1 – bc)), where k = (1 + bc)/{(1 + b2)(1 + c2)}.
(2.1)
Since L is the midpoint of PD we may obtain the co-ordinates of D as twice those of L minus those of P, the result being (x, y), where x = [2{(1 + bc)(b + c)} – p(1 + b2)(1 + c2)]/{(1 + b2)(1 + c2)}, (2.2) 2 2 y = [2(1 + bc)(1 – bc)]/ {(1 + b )(1 + c )}, (2.3) The equation of BC is (b + c)x + (1 – bc)y = (1 + bc). After some algebra the equation of the line perpendicular to BC passing through D is (1 – bc)x – (b + c)y + p(1 – bc) = 0. These lines meet at the midpoint of DX whose co-ordinates are (x, y), where x = (1/s)(– b2c2p + (1 + bc)(b + c) + 2bcp – p)), y = (1/s)((1 – bc)(bc + (b + c)p + 1)),
(2.4)
(2.5)
(2.6) (2.7)
where s = (1 + b2)(1 + c2). The co-ordinates of X follow and are found to be (x, y), where x = (p/s)(– b2c2 + b2 + c2 + 4bc – 1), (2.8) y = (2p/s)(b + c)(1 – bc). (2.9) The co-ordinates of Y, Z may be written by cyclic change of a, b, c. These co-ordinates may now be substituted into the general equation of a circle, x2 + y2 + 2gx + 2fy + t = 0, to provide three equations for f, g, t, which, with help from DERIVE, gives f = g = 0 and t = – p2, as required for the circle XYZ to have centre O and to pass through P. The equation of AP is (1 – a2)x + ((1 + a2)p – 2a)y – (1 – a2)p = 0. This meets the circle x2 + y2 = p2 again at the point U with co-ordinates (x, y), where x = (1/q)(p{a4(1 – p2) + 4a3p – 2a2(p2 + 3) + 4ap – p2 + 1}), y = (2/q)(p(1 – a2)(a2p – 2a + p)), and q = a4(1 + p2) – 4a3p + 2a2(1 + p2) – 4ap + p2 + 1. The co-ordinates of V and W may be written by replacing a by b and c respectively. 3
(2.10)
(2.11) (2.12) (2.13)
3.
The point Q and the indirect similarity
We now show that the lines UX, VY, WZ are concurrent at a point Q. The equation of the line UX turns out to be [a2{b(cp – 1) – c – p} + 2a(1 – bc) + b(1 + cp) + c – p]x + [a2{b(c + p) + cp – 1} – 2a(b + c) + b(p – c) + cp + 1]y + [a2{b(cp + 1) + c – p} + 2a(1 – bc) + b(cp – 1) – c – p]p = 0. (3.1) The equations of VY, WZ may be written down by cyclic change of a, b, c. These lines are concurrent at the point Q with co-ordinates (x, y), where x = (1/r){p(a2(b2(c2(p2 – 3) + p2 + 1) + 4b(c – p) + c2(p2 + 1) – 4cp + p2 + 1) + 4a(b2(c – p) + b(c2 – 2cp + 1) – c(cp – 1)) + b2(c2(p2 + 1) – 4cp + p2 + 1) + 4bc(1 – cp) + c2(p2 + 1) + p2 – 3)}, (3.2) 2 2 2 2 2 2 y = (2/r){p(a (b (c p – 2c + p) + 2bc(p – c) + p(c – 1)) – 2a(b c(c – p) + bp(1 – c ) + cp – 1) + 2 b p(c2 – 1) + 2b(1 – cp) – c2p + 2c – p)}, (3.3) and r = (1 – p2)(1 + a2)(1 + b2)(1 + c2). (3.4) Two triangles ABC and XYZ are similar when their angles are equal and this is so if their corresponding sides are in fixed ratio, that ratio being an enlargement (or reduction) about a given point. When they are indirectly similar it is always the case that there are two axes perpendicular to one another and passing through a fixed point such that when ABC is reflected in one of these axes through the fixed point and enlarged (or reduced) by a fixed amount through the fixed point, it is mapped on to triangle XYZ. The axes are called the double lines of inverse similarity and the fixed point is called the centre of inverse similarity. Either line of symmetry may be used, but in one case the enlargement factor is positive, and in the other case it is negative, implying that a rotation of 180o is also involved. What we prove now is that the triangles ABC and XYZ are indirectly similar, and we do this by establishing the double lines of inverse similarity and the centre of inverse similarity. The latter is, in fact, the circumcentre O. The axes are shown in Figure 2 as the lines m and n passing through O. In the figure the axis labelled m is used. The reflection through m takes triangle ABC into triangle A'B'C' and then there is a reduction (since P is inside ABC in the case we have drawn) by a factor OX/OA' taking A' to X. Since the line m turns out to be symmetric in a, b, c and OX/OA' = p, this shows that B' is taken to Y and C' to Z, thereby establishing the similarity.
4
m A
A'
P
X
N V
F
M
H O
E
E'
Q F' C'
Z
n W
T
C
Y
L
B
B'
U S
D' D
Figure 2
Suppose the equation of one of the axes of inverse symmetry passes through O and has equation y = mx. The equation of the line through A perpendicular to this line is (1 + a2)x + m(1 + a2)y + a2m – 2a – m = 0. (3.6) These lines meet at the point with co-ordinates (x, y) where x = (2a + m – a2m)/{(1 + a2)(1 + m2)}, y = m(2a + m – a2m)/{(1 + a2)(1 + m2)}.
(3.7) (3.8)
The co-ordinates of A', the reflection of A in y = mx, can now be obtained and are (x, y), where x = 2(m + a)(1 – am)/{(1 + a2)(1 + m2)}, (3.9) 2 2 y = ((a – 1) + m(a + 1))((a + 1) – m(a – 1))/{(1 + a )(1 + m )}. (3.10) We now determine the possible values of m if A', X, O are collinear. The condition for this is x1y2 = x2y1, where (x1, y1) are the co-ordinates of A' and (x2, y2) are the co-ordinates of X. This results in a quadratic equation for m whose solutions are, say, m and n, where mn = – 1 and 5
m = (abc + bc + ca + ab – a – b – c – 1)/(abc – bc – ca – ab – a – b – c + 1).
(3.11)
The enlargement (reduction factor) is obviously p (supposed positive without loss of generality). As the values of m and the enlargement factor are symmetric in terms of a, b, c and independent of them respectively, the proof of the indirect similarity is complete. 4.
Consequences and the connection with a circle through H of radius p
Now it is well known that a Hagge circle and its key points arise from an indirect similarity of the circumcircle, A, B, C and three other key points on the circumcircle, so it is a challenge to discover a direct similarity between the circle centre O through P, that we have so far been involved with, and a Hagge circle. This proves to be possible and depends on the remarkable properties of the point T defined to be the inverse image of Q in the indirect similarity described in Section 3. See Figure 3 for an illustration of what follows. Since T is the inverse image of Q, which we know is the point of concurrence of UX, VY, WZ, it follows that if we draw AT, BT, CT to meet the circumcircle at D', E', F' respectively, then not only is ABC indirectly similar to XYZ, but triangle D'E'F' is similar to triangle UVW. If therefore we create the Hagge circle generated by T, by reflecting D', E', F' in BC, CA, AB respectively to get points U', V', W', then triangle U'V'W' is both similar to triangle D'E'F', but then also to triangle UVW. Furthermore from the theory of Hagge circles it follows that if AH, BH, CH meet circle U'V'W' at X', Y', Z' respectively, then triangle X'Y'Z' is similar to triangle ABC and hence to triangle XYZ. Also from the theory of Hagge circles U'X', V'Y', W'Z' meet at T. This is part of what is needed to show that circle U'V'W'X'Y'Z' is directly similar to circle UVWXYZ. The second property of T is that it is the isogonal conjugate of P. This is best proved by drawing AT and AP to meet the circumcircle at a pair of points whose displacement vector is parallel to BC and similarly for the other vertices. We are now very much in business, because (again quoting the theory of Hagge circles) the figure POO'H must be a parallelogram, where O' is the centre of the Hagge circle. All is now clear, because the direct similarity between U'V'W'X'Y'Z' and UVWXYZ must be a 180o rotation about the midpoint R of OO'. The conclusive third property of T is that it is not only the inverse image of Q, but that R is the midpoint of QT.
6
m m' A
A'
X
P N H
V F F' C'
M
U'
Y'
E'
O
E Z'
n'
W' Z T O'
B
C
Y L
U
S
n
W
Q
R
V'
B'
X' D' D
Figure 3
The whole argument may be reversed. Create a Hagge circle, centre O', from a point T, perform an 180o rotation about the midpoint of OO' and you get a circle centre O of the same radius, and the remarkable thing is that this circle passes through P = Tg, the isogonal conjugate of T and, of course, TgOO'H is a parallelogram. It is also the case that R is the midpoint of UU', VV', WW', XX', YY', and ZZ'. Further features in Figure 3 are S, the point on the circumcircle where DD', EE', FF' appear to meet, the double lines of inverse symmetry m', n' of the indirect similarity between ABC and X'Y'Z', which are parallel to the axes m, n respectively. Indeed if one reflects ABC in n' to obtain triangle A''B''C'' it will be found that A''X', B''Y', C''Z' all pass through T. And finally the figure illustrates the fact that OQO'T is also a parallelogram. 7
What is needed in conclusion is a proof, which we now provide that the three properties leading to the position of T result in the same values for its co-ordinates. On the way we also find the coordinates of O'. The dilemma facing author and reader is that the algebra involved becomes technically very involved, even more so than what has preceded; so anyone checking the details will need an algebra computer package, such as DERIVE, which is the one we used. In logical order we first find the co-ordinates of the point T, when defined as the pre-image of the point Q in the indirect similarity. The co-ordinates of Q are given in Equations (3.3) and (3.4), but as these are very lengthy expressions, for the time being we call them (e, f). We want the reflection of (e, f) in the line y = mx, where m is given by Equation (3.11). The line perpendicular to y = mx through (e, f) has equation x + my = e + mf. This meets y = mx at the point ((e + mf)/(1 + m2), m(e + mf)/(1 + m2)). The reflection of Q has co-ordinates twice these minus those of Q. After dividing by p and inserting the values of e, f and m, this gives the following expressions for the co-ordinates of T, which are (x, y) where x = (2/r)(a2(b2(c2p + c(1 – p2) – p) – b(c2(p2 – 1) + 2cp – p2 – 1) – c2p + c(p2 + 1) – p) – a(b2(c2(p2 – 1) + 2cp – p2 – 1) + 2bp(c2 – 2cp + 1) – c2(p2 + 1) + 2cp + p2 – 1) – b2(c2p – c(p2 + 1) + p) + b(c2(p2 + 1) – 2cp – p2 + 1) – c2p + c(1 – p2) + p), (4.1) y = (1/r)(a2(b2(c2(p2 – 3) + 4cp – p2 – 1) + 4bcp(c – p) – (c2 – 1)(p2 + 1)) + 4ap(b2c(c – p) + bp(1 – c2) + cp – 1) + b2(1 – c2)(p2 + 1) + 4bp(cp – 1) + c2(p2 +1) – 4cp – p2 + 3), (4.2) and where r is given by Equation (3.4). Next we work out the isogonal conjugate of P with respect to triangle ABC and show that it coincides with T. The line AP has equation (a2 – 1)x – (a2p – 2a + p)y – p(a2 – 1) = 0. This meets the circumcircle at a point D'' with co-ordinates (2(a2p – a(p2 + 1) + p), (1 – p2)(a2 – 1)) / (a2 (p2 + 1) – 4ap + p2 + 1).
(4.3)
(4.4)
The line through D'' parallel to BC meets the circumcircle again at the point D'. When the coordinates of A, T and D' are entered into a determinant with 1 in the last column of each row, the value of this determinant is zero. Therefore the three points are collinear. Cyclic change shows that B, T, E' and C, T, F' are also collinear and hence T is the isogonal conjugate of P.
8
A
F' X
P
F
U'
O V Z
Q Y'
H Z'
Y R O'
U B
E
Pg
E'
V' C
W'
X'
D' S D
Figure 4
It follows that a Hagge circle generated by T can be drawn, using the points D', E', F' and if its centre is denoted by O', we know from the theory of Hagge circles that HTgOO' is a parallelogram. It follows, since Tg = P has co-ordinates (p, 0) that O' has co-ordinates (2a/(1 + a2) + 2b/(1 + b2) + 2c/(1 + c2) – p, (1 – a2)/(1 + a2) + (1 – b2) /(1 + b2) + (1 – c2)/(1 + c2)). (4.5) Our proposition that the direct similarity between the circle, centre O through P, and the Hagge circle, centre O', is correct, is confirmed if the 180o rotation of Q about R, the mid-point of OO', takes Q to T. And indeed the co-ordinates of O' minus the co-ordinates of Q do coincide with those of T, given in Equations (4.1) and (4.2). 9
5.
Further Observations
It is also the case that O' is the centre of circle DEF and that the isogonal conjugate of Q with respect to triangle DEF lies on the Hagge circle centre O'. The circle DEF is therefore directly similar to circle ABC and the circle centre O through P is the Hagge circle of Q with respect to triangle DEF. See Figure 4, where the full symmetry of the construction is finally revealed. We observe that DD', EE', FF' do concur at a point S on the circumcircle. Its co-ordinates are too complicated to record. Another rather curious result, as David Monk [3] pointed out, is that the centroid of triangle TPQ coincides with that of ABC. Using P to mean the vector OP etc., the proof of this is that, since T + Q = 2R = O', it follows that T + P + Q = O' + P = H = A + B + C. 6.
Conjecture supported by Cabri
We conclude with a conjecture. As we have seen circles centre O carry triangles that are directly similar to Hagge circles by 180o rotation about the midpoint of the line OO', where O' is the centre of the Hagge circle. We now conjecture that any circle and the triangles produced on it by a direct congruence of a Hagge circle and its triangles have additional properties that make the matter interesting and significant. We now refer to Figure 5 and the points have meanings as attached to this diagram. As can be seen the point of rotation of the Hagge circle R is chosen arbitrarily and the angle of rotation is also arbitrary. Cabri is so accurate that we have no doubt the conjectures we describe are true. We have not proved them as it seems that algebra is not the medium for doing so, and it seems unlikely that we could put through a proof algebraically anyway.
10
Angle of rotation = 84.6405888197 ° A F' F
C' X' V'
E' U
O
Z'
Z K
P'
P
H
O' U' B
Q
Y S
W'
R B'
E
V
X
C
Y' S' W A' D'
D
N
Figure 5
M
In the Figure the point P generates the Hagge circle in the usual way with U, V, W the reflections of D, E, F in the lines BC, CA, AB respectively, where D, E, F are the intersections of AP, BP, CP with the circumcircle. X, Y, Z are the points on AH. BH, CH lying on circle UVWH. Q is the centre of the Hagge circle and S is the opposite end of the diameter HQ. The direct congruence of the Hagge circle by rotation though R is the circle, centre O', with images U', V', W', X', Y', Z'. Also S' is the image of S. In the case of the circle centre O, the fourth vertex K of the parallelogram O'QHK is the isogonal conjugate of P. The conjecture is that this point K is now a conjugate of P in the sense that if AK, BK, CK meet the circumcircle of ABC at D', E', F' respectively then if L, M, N are the intersections of DD', EE', FF' with BC, CA, AB respectively, L, M, N are collinear. (In the case of the circles centre O the conjugate is the isogonal conjugate 11
and LMN is the line at infinity.) Finally if S' is the image of S under the direct congruence, then X', Y', Z' are now reflections of the point S' in axes (dotted in Figure 5), that make angles 86.405... o with lines parallel to the altitudes of triangle ABC. (That part of the conjecture is obvious.) In Figure 5 the axis through P of indirect similarity relating ABC and XYZ is shown. The conjugation and the role of the point S at the other end of the diameter to H in the Hagge circle are the matters that require further investigation, but which we do not intend to pursue further. References 1. K. Hagge, Zeitschrift für Math. Unterricht, 38 (1907) 257-269. (German) 2. Christopher Bradley and Geoff C. Smith, On a construction of Hagge, Forum Geometricorum (7) 231—247(2007). 3. D. Monk, private communication.
Flat 4 Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
12
ARTICLE 22 SIX POINTS ON A CIRCLE Christopher J Bradley
1. Introduction A construction arising from a triangle ABC and a variable point P is described. In general it leads to a conic passing through pairs of points on the sides of ABC, but for one, and only one, particular choice of P the conic is a circle, whatever the shape of the triangle. It is shown that the special choice of P is the image of a 180o rotation of the isotomic conjugate of the symmedian
1
point (also known as the third Brocard point or X76) about the midpoint of the two Brocard points. The circle bears distinct resemblance to the triplicate ratio circle. The general construction is to draw AP, BP, CP to meet BC, CA, AB at points R, S, T respectively. From R on BC draw lines parallel to BA and CA to meet CA and BA respectively at E' and F respectively. From S on CA draw lines parallel to CB and AB to meet AB and CB respectively at F' and D respectively. From T on AB draw lines parallel to AC and BC to meet BC and AC respectively at D' and E respectively. It then transpires that D', E', F' are the feet of Cevians through a point Q' and that D, E, F are the feet of Cevians through a point Q. A corollary of Carnot’s theorem then implies that D, E, F, D', E', F' lie on a conic. As stated above this conic is a circle if, and only if, P is chosen to be the image of a 180o rotation of the isotomic conjugate of the symmedian point about the midpoint of the two Brocard points. Subsidiary results are (i) that triangle DEF and D'E'F' have the same area, (ii) that if the triangle A'B'C' is constructed by extending the sides DF', FE and E'D' as shown in the Figure, then triangles ABC and A'B'C' are in perspective and (iii) if P is chosen at K, the symmedian point, then the points Q, Q' are the Brocard points and the points where AK, BK, CK meet BC, CA, AB respectively are the points where Brocard’s ellipse touches the sides of ABC. As far as (ii) is concerned there are in all eight such perspectives arising from interchange of primed and unprimed letters. A similar property holds for the triplicate ratio circle. Areal co-ordinates are used throughout and in dealing with circles results established in Bradley [1] are used without proof. 2. The construction Suppose that P has co-ordinates (l, m, n), then the point R where AP meets BC has co-ordinates (0, m, n). Lines through a point on BC parallel to AB have equation z – k(x + y + z) = 0 and for the particular line through R we must set k = n/(m + n). The line therefore has equation nx + ny – mz = 0 and this meets CA, y = 0, at the point E'(m, 0, n). Similarly the line through R parallel to CA meets AB at F(n, m, 0). The remaining points may now be obtained by cyclic change and have co-ordinates F'(l, n, 0), D'(0, m, l), D(0, l, n), E(l, 0, m). Note that AF/FB = CE'/E'A, BD/DC = AF'/F'B and CE/EA = BD'/D'C, equations that should be compared to those pertaining in the construction of the triplicate ratio circle. The equation of the conic S passing through the six points D, E, F, D', E', F' may now be obtained and is mnx2 + nly2 + lmz2 – (l2 + mn)yz – (m2 + nl)zx – (n2 + lm)xy = 0. (2.1)
2
The working is not given as the result may be checked by substitution. It may also be checked that AD', BE', CF' are concurrent at the Cevian point Q'(1/n, 1/l, 1/m) and that AD, BE, CF are concurrent at the Cevian point Q(1/m, 1/n, 1/l). Result (i) of the introduction is generally true, each of the triangles DEF, D'E'F' having area 2lmn[ABC]/{(m + n)(n + l)(l + m)}, (2.2) where [ABC] is the area of ABC. It is also easy to show that DQ divides EF in the same ratio as F'Q' divides D'E', that EQ divides FD in the same ratio as D'Q' divides E'F' and that FQ divides DE in the same ratio as E'Q' divides F'D'. Thus Q is situated in triangle DEF as Q' is situated in triangle F'E'D'. 3. The circle From Equation (2.1) and [1] the equations to be satisfied for S to be a circle are l2 + mn + nl + lm = a2, m2 + mn + nl + lm = b2, n2 + mn + nl + lm = c2. the solutions of which are proportional to l = – b2c2 + c2a2 + a2b2, m = b2c2 – c2a2 + a2b2, n = b2c2 + c2a2 – a2b2.
(3.1)
(3.2)
The x-co-ordinate of the centre of the circle may be calculated using the formula in [1] and is a2(a4(b4 – 4b2c2 + c4) – a2(b2 + c2)(b4 – 5b2c2 + c4) + b2c2(b4 – 4b2c2 + c4)). Note that the co-ordinates of P are proportional to (1/b2 + 1/c2 – 1/a2, 1/c2 + 1/a2 – 1/b2, 1/a2 + 1/b2 – 1/c2), the midpoint M of the two Brocard points is ½(1/b2 + 1/c2, 1/c2 + 1/a2, 1/a2 + 1/b2), the isotomic conjugate Kt of the symmedian point are (1/a2, 1/b2, 1/c2). These are normalized to the same amount and P = 2M – Kt, and hence P is the image of a 180o rotation of Kt about M. The equation of the line KtP is (x – y)/c2 + (z – x)/b2 + (y – z)/a2 = 0, and evidently this line contains the centroid G(1, 1, 1).
(3.3)
In fact, as David Monk informed me, P may alternatively be obtained from Kt by means of a homothety about G with ratio – 2. He also pointed out this means that Kt is the Brocard midpoint of the antimedial triangle.
3
It is also easy to show that DΩ divides EF in the same ratio as F'Ω' divides D'E', that EΩ divides FD in the same ratio as D'Ω' divides E'F' and that FΩ divides DE in the same ratio as E'Ω' divides F'D'. Thus Ω is situated in triangle DEF as Ω' is situated in triangle F'E'D'. 4. The perspectives Incredibly there are eight different perspectives, one each corresponding to the following hexagons F'FDD'EE', FF'DD'EE' and others formed by exchanging the roles of D and D' and of E and E'. Only the first of these is shown in the Figure. It corresponds to the ordering l > m > n or in the case of the circle of a > b > c. There are eight orderings of the values of l, m, n corresponding to the eight hexagons. If two of these magnitudes are equal, then one of the points is at a midpoint of BC, CA or AB, but it does not invalidate the analysis or the conclusion. If all three of l, m, n are equal the hexagon degenerates and in that case there is nothing of interest. We give the analysis only for the hexagon F'FDD'EE'. However, in the subsequent Figure we show the eight vertices of perspective, as the large squares, where three lines intersect,
4
The equation of F'E' is nx – ly – mz = 0.
(4.1)
mx – ny + lz = 0.
(4.2)
m2x + l2y – lmz = 0.
(4.3)
The equation of DF is The equation of ED' is
The co-ordinates of the vertices of the triangle formed by these three lines are A'(l(mn – l2), 2lm2, m(l2 + mn)), B'(2l2m, m(nl – m2), l(nl + m2), C'(l2 + mn, m2 + nl, n2 – lm). The three lines AA', BB', CC' meet at the point with co-ordinates (1/(m2 + nl), 1/(l2 + mn), 1/2lm), which is the required vertex of perspective, marked as J in the diagram. The Cabri diagram shows that the eight vertices of perspective themselves form figures that are in multiple perspective from the vertices and from points on the sides of ABC. Cabri also indicates that three of the vertices of perspective are collinear with the points Q and Q'.
5
5. When P lies at the symmedian point In this case l = a2, m = b2, n = c2 and Q and Q' become the Brocard points Ω and Ω' respectively. The Figure shows that the seven point circle may be constructed if anything in a simpler fashion than the standard method using the triplicate ratio circle. Besides the circumcentre O, the symmedian point K and the Brocard points Ω and Ω', the points L, M, N are the intersections of pairs of Cevians of the points Ω and Ω'. To be precise L = BE^CF', M = CF^AD' and N = AD^BE'. Analytic details are left to the reader who may check all these points lie on the seven point circle whose equation is b2c2x2 + c2a2y2 + a2b2 z2 = a4yz + b4zx + c4xy. (5.1)
6
Reference 1.
C.J.Bradley, The Algebra of Geometry, Highperception, Bath, 2007.
Flat 4, Terrill Court, 12-14 Apsley Road BRISTOL BS8 2SP.
7
The Symmedian point and the Polar Line C J Bradley ARTICLE 23
Y
Z
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S W
V K
M
B U
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L
X F
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1.
Introduction
To construct the figure above start with a triangle ABC and its symmedian point K. Next draw AK, BK, CK to meet the circumcircle Σ at L, M, N respectively. Then draw the tangents at L, M, N to Σ creating the triangle DEF, where D is the intersection of the tangents at M and N etc. Next draw the tangents at A, B, C to Σ creating the triangle RST, where R is the intersection of the tangents at B and C. The next step is to draw BF and CE to intersect at U, with V, W similarly defined. The following results now hold: (i) D, A, K, U, L, R are collinear, as are E, B, K, V, M, S and F, C, K, W, N, T; (ii) The six points D, E, F, R, S, T lie on a conic Γ; (iii)The following lines meet at X: TAS, NM, WV, BC, ELF and the tangents to Γ at D and R; the following lines meet at Y: RBT, LN, UW, CA, FMD and the tangents to Γ E and S; and the following lines meet at Z: SCR, LM, UV, AB, END and the 1
The Symmedian point and the Polar Line C J Bradley tangents to Γ at F and T. (iv) X, Y, Z are collinear and lie on the polar line of K with respect to both Σ and Γ. Note also that there are two triangles ABC and LMN inscribed in Σ and two triangles RST and DEF inscribed in Γ. Two related porisms are thus created, the first being the Brocard porism with a circle as circumconic and the Brocard ellipse (not drawn) as inconic, the second being a related Brocard porism with the circle Σ as inconic and the conic Γ as circumconic. K is the symmedian point of the triangles in the first porism, and Gergonne’s point in the second porism. Lines through K drawn from a vertex of any triangle in either porism meet the opposite side of that triangle at the point of contact of that side with the inconic. See the figure in which L, M, N are such points for triangles in the second porism. See Bradley and Smith [1] for an account of the Brocard porism. In subsequent sections we prove these results using areal co-ordinates throughout. For a description of these and how to use these see Bradley [2]. 2. The points L, M, N We take ABC to be triangle of reference. K has co-ordinates (a2, b2, c2). The equation of AK is b2z = c2y. (2.1) This meets the circumcircle Σ, with equation a2yz + b2zx + c2xy = 0,
(2.2)
at the point L with co-ordinates (–a2, 2b2, 2c2). Similarly M has co-ordinates (2a2, – b2, 2c2) and N has co-ordinates (2a2, 2b2, – c2). 3. The points R, S, T, D, E, F The equation of the tangent at A is b2z + c2y = 0.
(3.1)
and the equations of the tangents at B and C may be obtained by cyclic change of a, b, c and x, y, z. The tangents at B and C meet at the point R with co-ordinates (–a2, b2, c2). Similarly S has co-ordinates (a2, – b2, c2) and T has co-ordinates (a2, b2, – c2). Note that R lies on AK with Equation (1). Similarly S, T lie on BK, CK respectively. The equation of the tangent at L (–a2, 2b2, 2c2) is 4x/a2 + y2/b2 + z/c2 = 0.
(3.2)
The tangents at M, N follow by cyclic change of a, b, c and x, y, z. These two tangents meet at the point D with co-ordinates (5a2, – b2, – c2). Similarly the point E has co2
The Symmedian point and the Polar Line C J Bradley
ordinates (– a2, 5b2, – c2) and the point F has co-ordinates (– a2, – b2, 5c2). Note that D, E, F lie on AK, BK, CK respectively. To determine the equation of the conic Γ passing through the six points D, E, F, R, S, T one forms a 6 x 6 determinant with x2, y2, z2, yz, zx, xy as the elements of row 1 and then in the next 5 rows one puts the values of these entries for 5 of the points. On putting this determinant equal to zero, the equation of the conic through the 5 points emerges. One may then check that it passes through the 6th point. In this case we find that Γ has equation x2/a4 + y2/b4 + z2/c4 + 3yz/(b2c2) + 3zx/(c2a2) + 3xy/(a2b2) = 0. 4.
(3.3)
The points X, Y, Z, the polar line and the line MN
The line EF has Equation (3.2), as it coincides with the tangent at L. Similarly the line TS has Equation (3.1), as it coincides with the tangent at A. We define X as the intersection of these two lines, which therefore has co-ordinates (0, b2, – c2). Similarly Y has co-ordinates (– a2, 0, c2) and Z has co-ordinates (a2, – b2, 0). It is clear that the line XYZ has equation x/a2 + y/b2 + z/c2 = 0.
(4.1)
Working out the polar of a point with respect to a conic involves the same algebra as working out a tangent, so it is now possible to check that XYZ is the polar of K with respect to both Σ and Γ. The co-ordinates of M and N are given in Section 2. The equation of the line MN is x/a2 = 2y/b2 + 2z/c2.
(4.2)
Clearly MN passes through X. 5.
The points U, V, W and the lines through X
The equation of the line EC is 5b2x + a2y = 0,
(5.1)
5c2x + a2z = 0.
(5.2)
and the equation of the line FB is These two lines meet at the point U with co-ordinates (– a2, 5b2, 5c2). Similarly V and W have co-ordinates (5a2, – b2, 5c2) and (5a2, 5b2, – c2) respectively. The line VW therefore has equation 4x/a2 = 5y/b2 + 5z/c2.
3
(5.3)
The Symmedian point and the Polar Line C J Bradley Again it is clear that VW passes through X. Note also that the tangent at L, with Equation (3.2), also passes through X. Finally the tangent at D to Γ has equation 2x/a2 + 5y/b2 + 5z/c2 = 0.
(5.4)
This line evidently passes through X. Since D and R lie on a line through K and the tangent at D passes through X, it follows that the tangent at R also passes through X.
References 1. C. J. Bradley & G. C. Smith, accepted for publication. 2. C. J. Bradley, Challenges in Geometry, Oxford, 2005.
Flat 4 Terrill Court 12-14 Apsley Road BRISTOL BS8 2SP
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ARTICLE 24 The Thirteen point Circle Christopher J Bradley
A1 A
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N2 M2 M L2 L K O M1 ' N1 N L
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C A2
1. Introduction In this article we give an account of the properties of the coaxal system of circles passing through the two Brocard points Ω and Ω' and having the Brocard axis as line of centres. This coaxal system contains four circles of particular importance, the seven point circle S and the three circles AΩΩ', BΩΩ' and CΩΩ', where A, B and C are the vertices of the triangle. The circumcircle is not a member of this coaxal system, nor is the coaxal system orthogonal to the coaxal system containing the Apollonius circles. We now introduce our notation. The centres of the four circles just mentioned are (i) the midpoint of OK, where O is the circumcentre and K the symmedian point, (ii) P, (iii) Q, and (iv) R. The circle AΩΩ' meets AB at A1 and CA at A2. The circle BΩΩ' meets BC at B1 and AB at B2. The circle CΩΩ' meets CA at C1 and BC at C2. A1Ω meets S at N1, B1Ω meets S at L1, C1Ω 1
meets S at M1, A2Ω' meets S at M2, B2Ω' meets S at N2 and C2Ω' meets S at L2. These additional points on S convert it from being the seven point circle to the thirteen point circle of the title. We establish the following results: (i) P, Q, R lie on the Brocard axis; (ii) BCB1 C2 meets B2C1 at a point A3 on the Brocard axis, CAC1A2 meets C2A1 at a point B3 on the Brocard axis and ABA1B2 meets A2B1 at a point C3 on the Brocard axis; (iii) A1B1C1 is a straight line passing through Ω' and A2B2C2 is a straight line passing through Ω and these two lines intersect at K, the symmedian point; (iv) M2Ω and N1Ω' pass through A3, N2Ω and L1Ω' pass through B3, and L2Ω and M1Ω' pass through C3. These results are illustrated in the Figure generated by CABRI software and the computer algebra package DERIVE was used to check the algebra for which areal co-ordinates are used throughout. 2. The seven point circle and the Brocard axis The Brocard axis passes through O and K and putting (x, y, z) and their co-ordinates as rows of a determinant we find its equation to be =0
(2.1)
In areal co-ordinates the equation of any circle may be put in the form ,
(2.2)
where u, v, w are constants to be determined, see Bradley [1, 2]. Putting in the co-ordinates of K(a2, b2. c2), Ω(1/b2, 1/c2, 1/a2), Ω'(1/c2, 1/a2, 1/b2) we obtain three equations to determine u, v, w and after some simplifications we obtain the equation of the seven point circle in the form . (2.3) 3. The three circles and their intersections with the sides of the triangle We obtain the equations of the circles AΩΩ', BΩΩ' and CΩΩ' using the same method and their equations turn out to be = 0,
(3.1) (3 .2)
(3.3) 2
respectively. The intersections of these circles with the sides of ABC are now easily found and are: A1(a2(a2 – b2), b2(a2 – c2) , 0), A2(a2(a2 – c2), 0, c2(a2 – b2)), B1(0, b2(b2 – c2), c2(b2 – a2)), B2(a2(b2 – c2), b2(b2 – a2), 0), C1(a2(c2 – b2), 0, c2(c2 – a2)), C2(0, b2(c2 – a2), c2(c2 – b2)) Since the perpendicular bisector of ΩΩ' is the Brocard axis it is clear that the centres P, Q, R of the circles AΩΩ', BΩΩ' and CΩΩ' all lie on the Brocard axis. 4. The lines A1B1C1 and A2B2C2 Using the usual determinantal method the equation of the line A1B1C1 is (4.1) and it may be checked that K(a , b , c ) and Ω'(1/c , 1/a , 1/b ) both lie on this line. Similarly the equation of the line A2B2C2 is (4.2) 2 2 2 2 2 2 And it may be checked that K(a , b , c ) and Ω(1/b , 1/c , 1/a ) both lie on this line. Note that the two lines intersect at K on the Brocard axis. 2
2
2
2
2
2
5. The points A3, B3, C3 The equation of C1B2 is ,
(5.1)
and the equation of B1C2 is x = 0. These meet at the point A3(0, b2(a2 – b2), c2(a2 – c2)), a point that clearly lies on the Brocard axis with Equation (2.1). Similarly B3 and C3 have co-ordinates B3(a2(b2 – a2), 0, c2(b2 – c2)) and C3(a2(c2 – a2), b2(c2 – b2), 0) both of which also lie on the Brocard axis. 6. The six points L1, L2, M1, M2, N1, N2 For the record L = BΩ^CΩ', M = CΩ^ AΩ', N = AΩ^BΩ' have co-ordinates L(a2, c2, b2), M(c2, b2, a2), N(b2, a2, c2) and together with O, K, Ω, Ω' form the seven points of the seven point (Brocard) circle. As stated in Section 1 the six additional points to provide the thirteen in the Figure are such that A1Ω meets S at N1, B1Ω meets S at L1, C1Ω meets S at M1, A2Ω' meets S at M2, B2Ω' meets S at N2 and C2Ω' meets S at L2. After some algebra we find their co-ordinates to be L1(a2b2, b2(c2 + a2 – b2), b4 + (a2 – b2)(c2 + a2)), M1(c4 + (b2 – c2)(a2 + b2), b2c2, c2(a2+ b2 – c2)), N1(a2(b2 + c2 – a2), a4 + (c2 – a2)(b2 + c2), c2a2), L2(c2a2, a4 + (a2 – c2)(b2 – c2), c2(a2 + b2 – c2)), M2(a2(b2 + c2 – a2), a2b2, b4 + (b2 – a2)(c2 – a2)), N2(c4 + (c2 – b2)(a2 – b2), b2(c2 + a2 – b2), b2c2).
3
7. More lines through A3, B3, C3 We give the analysis to show that M2Ω^N1Ω' = A3, the pairs of lines that intersect in B3 and C3 then follow by cyclic change of letters A, B, C and L, M, N. The equation of M2Ω is (7.1) The equation of N1Ω' is (7.2) These two lines intersect at A3(0, b2(a2 – b2), c2(a2 – c2)).
References 1. C. J. Bradley, Challenges in Geometry, Oxford, 2005. 2. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, 2007.
Flat 4 Terrill Court 12-14 Apsley Road BRISTOL BS8 2SP
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Article 25 When quadrangles are completely in perspective Christopher J Bradley 1. Introduction In this article we establish a sufficient condition for when a pair of quadrangles have what may be appropriately called a Desargues’ axis of perspective. It is also shown how the condition is automatically satisfied when corresponding pairs of vertices of the two quadrangles lie on a conic and are in involution by means of perspectivity. The ideas are then formulated and outlined to provide a guide as to what happens for polygons with an arbitrary number of sides. The analysis is carried out using projective co-ordinates. EF^UV CD^RS FG^VW AC^PR AD^PS EG^UW
F
W
BC^QR
A
R
Q
U
D
S
To BD^QS
To AB^PQ
X
E
B C P
To V
Figure 1
We define two types of perspective in dealing with the polygons A1A2…An and B1B2…Bn. Partial perspective is defined to mean that A1B1, A2B2, ..., AnBn are concurrent at a point X. Complete perspective is defined to mean that all the points AjAk^BjBk, j, k = 1 to n, j k are collinear. So complete perspective is when an axis exists that may be termed a Desargues’ axis of perspective.
V
1
G
2. Triangles Theorem 1 (Desargues’ theorem) When n = 3 and we are dealing with triangles, then Partial perspective perspective.
Complete
Proofs are given in all good geometry books, such as Pedoe [1].
3. Sufficient condition for quadrangles Theorem 2 Let ABCD and PQRS be in partial perspective with vertex X, so that P lies on AX, Q on BX, R on CX, and S on DX. If, in addition, AC^BD, PR^QS are collinear with X, then ABCD and PQRS are in complete perspective. What the extra condition means is that if one corresponding pair of diagonal points E = AC^BD and U = PR^QS are such that EU also passes through X, and if ABCD and PQRS are also in direct perspective with vertex X, then they are in complete perspective. Proof In the real projective plane we take a co-ordinate system such that A is (– 1, 1, 1), B is (1, – 1, 1), C is (1, 1, – 1), D is (1, 1, 1); the diagonal point triangle is then the triangle of reference and in particular E = AC^BD is (0, 1, 0). Let X be (f, g, h). Then P has coordinates (f – , g + , h + ) for some value of , Q has co-ordinates (f + , g – , h + ) for some value of , R has co-ordinates (f + , g + , h – ) for some value of and S has co-ordinates (f – , g – , h – ) for some value of . For the intersection of PR with EX we want a linear combination of P and R satisfying hx = fz. This is easily found to be P + R = (f( + ), g( + ) + 2 , h( + )). Similarly the intersection of QS with EX is Q + S = (f( + ), g( + ) + 2 , h( + )). We require these to be the same point, the condition for which is 2 /( + ) = 2 /( + ) or + + + = 0. (3.1) Consider now the axis of perspective of triangles ABC and PQR. One point on it is the intersection of AB (x + y = 0) and PQ. This is P – Q = (– – , + , – ). Another is the intersection of BC (y + z = 0) and QR, which is Q – R = ( – , – – , + ). The line joining these is easily calculated to be ( + )x + ( + )y + ( + )z = 0. (3.2) 2
From Desargues theorem for the triangle, CA^RP must lie on this line. We must show that AD^PS also lies on this line. AD has equation y = z and the intersection is P – S = ( – , + , + ). So the condition, from equation (2), is ( + )( – ) + ( + )( + ) + ( + )( + ) = 0, and this reduces to Equation (1). Similarly BD^QS, CD^RS lie on the axis of perspective.
Theorem 3 Under the same conditions as Theorem 2, the diagonal point triangles EFG of ABCD and UVW of PQRS are in perspective, with vertex of perspective X and with the same axis of perspective as the quadrangles themselves. Proof The co-ordinates of F are (0, 0, 1). By working analogous to that in Theorem 2, equation (2) ensures that the intersection of PQ and FX is the same point as the intersection of RS with FX, and this shows that AB^CD, X and PQ^RS are collinear. Similarly AD^BC, X and PS^QR are collinear. This shows that triangles EFG and UVW are in perspective with vertex X. It is straightforward, though tedious, to show that the axis of perspective of the two diagonal point triangles is the same axis of perspective as the quadrangles themselves and a proof is omitted.
Figure 1 illustrates Theorems 2 and 3. An alternative theorem to Theorem 2 is possible. If one starts with triangles ABC and PQR in perspective, so that what will become the Desargues’ axis is located. Then D can be placed anywhere and the position of S may be fixed, not by the conditions of Theorem 2, but by requiring AD^PS and BD^QS both lie on the Desargues’ axis. It then follows that CD^RS automatically lies on the Desargues’ axis and the quadrangles are in complete perspective, as before. It follows, since D, S, X are then collinear, that for quadrangles complete perspective implies direct perspective. 4. Polygons It is now possible to see how to generalise the result for quadrangles to polygons with an arbitrary number of sides. As before direct perspective plus a number of other conditions ensure complete perspective. To get the idea let us consider pentagons ABCDE and PQRST. Let AP, BQ, CR, DS, ET be concurrent at X. Fix D and S as in Theorem 2 so that quadrangles ABCD and PQRS are in complete perspective. Then E and T are selected in similar fashion so that AC^BE, X, PR^QT are collinear. Then it follows that ABCDE and PQRST are in complete perspective. Essentially one builds up the pentagon from the two pairs of complete quadrangles ABCD, PQRS and ABCE, PQRT. Admittedly this is merely a guide to what happens and does not constitute a proof, though the figure that follows is convincing. 3
Figure 2 Figure 2 shows two pentagons in complete perspective. It shows how the points S and T are selected in order to make 1, X, 1 collinear, where 1 = AC^BD and 1' = PR^QS and 2, X, 2 collinear, where 2 = AC^BE and 2' = PR^QT. Note that when this is done it is automatic that DE^ST lies on the Desargues’ axis. 5. Quadrangles inscribed in a conic We now consider a special case when two quadrangles in direct perspective are automatically in complete perspective. Theorem 4 Let ABCD and PQRS be two distinct quadrangles in direct perspective with vertex Y, and suppose all the vertices lie on a conic not passing through Y, and then the quadrangles are in complete perspective. Co-ordinate Proof We show that AC^BD, Y, PR^QS are collinear, so that Theorem 2 applies. Let the conic have equation y2 = zx, with points U on having co-ordinates given parametrically as (u2, u, 1). Then the equation of UV is x + uvz = (u + v)y. Suppose 4
now that Y has co-ordinates (0, 1, 0), where Y does not lie on , then points U, V at opposite ends of chords through Y are in involution by means of the perspectivity through Y, given analytically by v = – u. Let A, B, C, D have parameters a, b, c, d, then P, Q, R, S have parameters – a, – b, – c, – d. The equation of AC is x – (a + c)y + acz = 0 and that of BD is x – (b + d)y + bdz = 0 and AC^BD has co-ordinates (ac(b + d) – bd(a + c), ac – bd, a + c – b – d). Similarly PR^QS has co-ordinates (– ac(b + d) + bd(a + c), ac – bd, – a – c + b + d). Their collinearity with Y is now immediate. Two points need mentioning. The first is that in the real projective plane Y is external to the conic in that not all lines through Y intersect the conic in real points. However, in the complex plane all lines through Y intersect the conic in two points. So it is perfectly general to take Y as a general point of the plane. The second observation is that if ac = bd then the two points above coincide. This is perfectly possible, but does not invalidate the theorem as we may then take two other corresponding diagonal points, and not all pairs of diagonal points can coincide unless the quadrangles coincide, which we have assumed not to be the case. Synthetic proof From the inscribed quadrangle ADPS the intersection AD^PS lies on the polar of Y and similarly for the other five intersections.
Since A and P may be interchanged and also B and Q, C and R, and D and S it is also the case that such intersections as AB^PQ, AC^PR, BC^QR, BD^QS, CD^RS also lie on the polar of Y. For the purposes of clarity and because of available space only a few of the intersections are shown in Figure 3.
5
Figure 3
Theorem 5 In the same configuration as Theorem 4, let AYP meet BD and QS at A and P , let CYR meet BD and QS at C and R , let BYQ meet AC and PR at B and Q and let DYS meet AC and PR at D and S . Then A , B , C , D , P , Q , R , S lie on a conic and the polar of Y with respect to this conic is the same Desargues’ axis as for the quadrangles ABCD, PQRS. Proof The equation of BD is x – (b + d)y + bdz = 0 and the equation of AYP is x = a2z and these meet at A , with co-ordinates (a2(b + d), a2 + bd, (b + d)). Likewise P has coordinates (– a2(b + d), a2 + bd, – (b + d)). The co-ordinates of C and R follow by 6
replacing a2 by c2. The other four points now follow with a, c replacing b, d and b2, d2 replacing a2, c2. The equation of the conic that passes through all eight points may now be verified to be ((a+ c)2 + (b + d)2)x2 – (a + c)2(b + d)2y2 + (b2c2d2 + a2c2d2 + a2b2d2 + a2b2c2 + 2abcd(ac + bd))z2 – ((a2 + c2)(b2 + d2) – 4abcd)zx = 0. (5.1) Furthermore the fact that xy and yz terms in equation (3) are absent means that the polar of Y with respect to this conic has equation y = 0, which is also the polar of Y with respect to the original conic y2 = zx.
Corollary It now follows that the quadrangles A B C D and P Q R S are in perspective with vertex Y and since they lie on a conic are in complete perspective. Thus intersections such as A B ^P Q also lie on the same Desargues’ axis of perspective. See Figure 3 as illustration of this. It is also the case that polygons of an arbitrary number of sides that are in direct perspective and whose vertices lie on a conic are in complete perspective, though the proof is beyond the intention of this article. 6. Cross ratio Consider the quadrangle ABCD inscribed in the conic with equation y2 = zx, where A, B, C, D have parameters a, b, c, d then {A, C; B, D} = {a, c; b, d}. If the quadrangle with vertices P, Q, R, S is defined to be in perspective through Y, as in Theorems 4 and 5, then its parameters are – a, – b, – c, – d respectively and since {– a, – c; – b, – d} = {a, c; b, d}, it follows that {A, C; B, D} = {P, R; Q, S}. Since the point Y is at our disposal we have proved the following theorem: Theorem 6 Two quadrangles ABCD and PQRS in direct perspective on a conic are such that {A, C; B, D} = {P, R; Q, S}.
Reference 1. Dan Pedoe, Geometry, A Comprehensive Course, Cambridge, 1970.
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Article 26 A circle concentric with the incircle Christopher J Bradley
Figure 1 1. Introduction Let ABC be a triangle and P a variable point. Draw AP, BP, CP to meet BC, CA, AB respectively in U, V, W. Draw the circles UPV, VPW, WPU. These meet the sides in 6 points labelled Q, R, S, T, E, F in the diagram.
1
Then it turns out that these 6 points always lie on a conic. Furthermore, this conic is a circle if, and only if, P is Gergonne’s point for ABC, and then the centre of the circle is the incentre I. Now let L, M, N be the centres of circles VPW, WPU, UPV respectively. We prove that triangles ABC and LMN are in perspective with vertex of perspective I. These properties are displayed in Figure 1. We prove also that the perspector (the Desargues’s axis of perspective) is parallel to the polar of Gergonne’s point with respect to the incircle. In the proof of these results we use areal coordinates, a description of whose definition and properties may be found in Bradley [1]. 2. The six points Take P to have co-ordinates (l, m, n), then U, V, W have co-ordinates (0, m, n), (l, 0, n) and (l, m, 0) respectively. Suppose the equation of the circle VPW is a2yz + b2zx + c2 xy – (x + y + z)(ux + vy + wz) = 0. (2.1) As shown in [1] all circles have an equation of this form for variable u, v, w. Putting in the coordinates of V, P, W we can now find the values of u, v, w particular for this circle. They are u = {mn(l(b2(l + m) + c2(l + n)) – a2(l + m)(l + n))}/{l(l + m)(l + n)(l + m + n)}, (2.2) 2 2 2 v = {a n(l + n) – l(b n – c (l + n))}/{(l + n)(l + m + n)}, (2.3) 2 2 2 w = {a m(l + m) + l(b (l +m) – c m)}/{(l + m)(l + m + n)}. (2.4) Equations (2.2) to (2.4) may now be substituted in equation (2.1) to obtain the equation of circle VPW, which is mnx2(a2(l + m)(l + n) – l(b2(l + m) + c2(l + n))) – ly2(l + m)(a2n(l + n) – l(b2n – c2(l + n))) + lz2(l + n)(a2m(l + m) + l(b2(l + m) – c2m)) + l2yz(l(b2(l + m) + c2(l + n)) – a2(l + m)(l + n)) – xy(a2n(l + m)(l – m)(l + n) – l(b2n(l + m)(l – m) + c2(l + n)(l(m + n) + m2))) + zx(a2m(l + m)(l + n)(l – n) – l(b2(l + m)(l(m + n) + n2) + c2m(l + n)(l – n))) = 0. (2.5) This circle meets CA at the point Q with co-ordinates ((l + n)(a2m(l + m) + l(b2(l + m) – c2m)), 0, m(l(b2(l + m) + c2(l + n)) – a2(l + m)(l + n))). From similar working the point R where this circle meets AB has co-ordinates ((l + m)(l(b2n – c2(l + n)) – a2n( l + n)), n(a2(l + m)(l + n) – l(b2(l + m) + c2(l + n))), 0). The hard way of proceeding from here would be to obtain the co-ordinates of the four points S, T, E, F by cyclic change of letters and then show that all six points lie on a circle if, and only, if (l, m, n) are the co-ordinates of Gergonne’s point. It is easy to see that the analytic work involved in proceeding in this way would be very difficult. The reason it was possible to be wise before the event is that by good fortune we discovered that P was Gergonne’s point by guesswork, using CABRI software. That all six points lie on a conic whatever the choice of P made it likely that 2
some choice of the position of P, and no other, would result in the conic being a circle. This allows the following easier approach. The values of l, m, n for P to be Gergonne’s point are l = (c + a – b)(a + b – c), m = (a + b – c)(b + c – a), n = (b + c – a)(c + a – b).
(2.6)
With these values the co-ordinates of Q are ((a + b – c)(2a2 – a(4b + c) + 2b2 – bc – c2), 0, (b + c – a)(2a2 – a(b + c) – (b – c)2), and by cyclic change of letters the circle WPU meets AB at the point T with co-ordinates ((b – c – a)(a2 + a(b – 2c) – 2b2 + bc + c2), (a – b – c)(a2 + a (b + c) – 2(b2 – 2bc + c2)), 0) and the circle UPV meets BC at E with co-ordinates (0, (c – a – b)(a2 + a(c – 2b) + b2 + bc – 2c2), (a – b + c)(2a2 – a(b + 4c) – b2 – bc + 2c2)). It is now possible to find the equation of the circle Σ containing the points Q, T and E. There is no need to give the working as the reader may verify that the three points with co-ordinates just given lie on the conic whose equation we now give, and also to check, by means of Section 2.3.10 of [1] that this conic is a circle. The equation is (b + c – a)2(2b2 + 2c2 – a2 – 4bc – ab – ca)(b2+ c2 – 2a2 – 2bc + ab + ca)x2 + … + … + (a + b – c)(c + a – b){5a4 – 11a3(b + c) + a2(3b2 + 3c2 + 20bc) + 7a(b3 – b2c – bc2 + c3) – 4b4 – 2b3c + 12b2c2 – 2bc3 – 4c4}yz + … + … . = 0 (2.7) Here the terms in y2 and z2 follow from those of x2 by cyclic change of a, b, c. Also the terms in zx, xy follow from those of yz by means of cyclic change of a, b, c. The point R has co-ordinates ((b – c – a)(2a2 – a(b + 4c) – b2 – bc + 2c2), (a – b – c)(2a2 – a(b + c) – (b – c)2), 0) and it may now be checked that R lies on the circle Σ with Equation (2.7). Similarly, by symmetry, S and F lie on Σ, which therefore contains all six points Q, R, S, T, E, F. 3. The circle VPW and its centre L The circle VPW, for P having general co-ordinates (l, m, n), has Equation (2.5). Putting the values (2.6) for Gergonne’s point into Equation (2.5) we obtain its equation as ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (3.1) where u = – (b + c – a)2(2a2 – a(b + c) – (b – c)2), v = – (c + a – b)2(2a2 – a(b + 4c) – b2 – bc + 2c2), w = – (a + b – c)2(2a2 – a(4b + c) + 2b2 – bc – c2), 2f = (a + b – c)(c + a – b)(– 2a2 + a(b + c) + (b – c)2), 2g = – (b + c – a)(a + b – c)(– 4a2 + a(5b + 2c) – b2 – bc + 2c2), 2h = (c + a – b)(b + c – a)(4a2 – a(2b + 5c) – 2b2 + bc + c2).
3
Using Equation (2.9) of Section (2.3.6) of [1], we find the co-ordinates of the centre L of this circle to be L(2a3 – 7a2(b + c) + 2a(b – c)2 + 3(b + c)(b2 – 2bc + c2), b(5a2 – 4a(b + c) – (b – c)2), c(5a2 – 4a(b + c) – (b – c)2)). (3. 2) It is immediate that AL passes through I(a, b, c). Similarly BM and CN pass through I and hence triangles ABC and LMN are in perspective with perspector the incentre of ABC. The coordinates of M and N follow from cyclic change of a, b, c and x, y, z and are hence M(a(5b2 – 4b(c + a) – (c – a)2), 2b3 – 7b2(c + a) + 2b(c – a)2 + 3(c + a)(c2 – 2ca + a2), c(5b2 – 4b(c + a) – (c – a)2)), (3. 3) 2 2 2 2 3 2 2 N(a(5c – 4c(a + b) – (a – b) ), b(5c – 4c(a + b) – (a – b) ), 2c – 7c (a + b) + 2c(a – b) + 3(a + b)(a2 – 2ab + b2)). (3.4) 4. The perspector is parallel to the polar of Ge with respect to the incircle The equation of the incircle of ABC is known to be (b + c – a)2 x2 + (c + a – b)2y2 + (a + b – c)2z2 – 2(a + b – c)(c + a – b)yz – 2(b + c – a)(a + b – c)zx – 2(c + a – b)(b + c – a)xy = 0. (4.1) The equation of the polar of Gergonne’s point with respect to this circle is easily shown to be (b + c – a)x + (c + a – b)y + (a + b – c)z = 0. (4.2) It is a straightforward but technically difficult piece of analysis to find the equation of MN and its intersection U' with the line BC. U' is, of course, on the perspector. Its co-ordinates are (0, (a + b – c)(–a2 + a(2b – 4c) – b2 – 4bc + 5c2), (c + a – b)(a2 + a(4b – 2c) – 5b2 + 4bc + c2)) (4.3) Lines parallel to the polar of Gergonne’s point, with equation (4.2), have equations of the form (b + c – a)x + (c + a – b)y + (a + b – c)z + k(x + y + z) = 0. (4.4) The line with Equation (4.4) contains U' if, and only if, k = 3(a3 – a2(b + c) – a(b – c)2 + (b + c)(b – c)2)/{2(a2 – 2a(b + c) + (b – c)2)}.
(4.5)
Inserting the value of k from (4.5) into Equation (4.4) we obtain the line parallel to the polar of Gergonne’s point through U' to have equation (b + c – a)(5a2 – 4a(b + c) – (b – c)2)x + (c + a – b)(5b2 – 4b(c + a) – (c – a)2)y + (a + b – c)(5c2 – 4c(a + b) – (a – b)2)z = 0. (4.6) The symmetry of this equation under the cyclic permutations a, b, c and x, y, z shows that it not only contains U' = BC^MN, but also V' = CA^NL and W' = AB^LM and is therefore the perspector. It is left as an exercise to show that this line passes through the midpoint of Ge and the inverse of Ge with respect to the incircle. 4
Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath (2007).
Flat 4, Terrill Court 12-14, Apsley Road BRISTOL BS8 2SP
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Porisms with a circular circumconic C J Bradley and G C Smith 1. Introduction
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1. Introduction This article is concerned with triangle porisms. Whenever there is a single triangle that circumscribes an inconic and is inscribed in a circumconic, it is known that there are an infinite number of such triangles. See, for example, Maxwell [1]. We are concerned here with porisms in which the circumconic is the circumcircle of all the triangles involved. If the inconic intersects the circumcircle we call such porism an incomplete porism and there are points on the circumconic that cannot entertain a triangle vertex. Those in which a vertex may appear at any point of the circumcircle we call a complete porism. The complete porisms are those encountered most frequently and include Poncelet’s porism, when the inconic is the incircle, and Brocard’s porism, when the inconic is Brocard’s ellipse. See Bradley and Smith [2] for an account of this porism, which has the remarkable property that any two triangles in the porism are in triple reverse perspective. The result we prove in this article is if we take the orthocentre of all the triangles in a case when the circumconic is the circumcircle, then the path traced out by the orthocentre is circular (or linear). For a complete porism it is an entire circle, but for an incomplete porism it is a circular arc (or segment). It is when the inconic is a parabola that the locus is part of a straight line. Moreover we are able to identify the circle as the common Hagge circle [3] of the triangles involved or in the case of an incomplete porism part of such a circle. Analysis is carried out using areal co-ordinates. In what follows we refer repeatedly to Figure 1, introducing points on the diagram as they occur in the text. 2. Key points of the analysis We start with a triangle ABC as triangle of reference and its circumcircle, which we refer to as Γ. The orthocentre is H(u, v, w), where u : v : w = tan A : tan B : tan C so that the equation of the circumcircle is u(v + w)yz + v(w + u)zx + w(u + v)xy = 0. (2.1) The co-ordinates of O, the centre of Γ are (v + w, w + u, u + v) and the co-ordinates of the ninepoint centre are (2u + v + w, 2v + w + u, 2w + u + v). A parameter system on Γ may be set up so that a point R, with parameter r, has co-ordinates (– u(v + w)r(1 – r), v(w + u)(1 – r), w(u + v)r), as may be checked by substitution in Equation (2.1). The porism is created by choosing a point I to act as the centre of an inconic. (Note it can be inscribed or escribed.) We take the equation of the inconic to be l2x2 + m2y2 + n2z2 – 2mnyz – 2nlzx – 2lmxy = 0. (2.2) This meets the side BC, x = 0, where (my – nz)2= 0, the double root indicating tangency, and the point of tangency being (0, 1/m, 1/n). The Brianchon point (the Cevian point for which the feet of the Cevians are the points of tangency) for this inconic has co-ordinates (1/l, 1/m, 1/n).The
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polar of the point with co-ordinates (m + n, n + l, l + m) with respect to the inconic is x + y + z = 0, the line at infinity, showing the co-ordinates of I are (m + n, n + l, l + m). We introduce a point h lying on OI such that OI = Ih. The normalized x-co-ordinate of h is given by (2Ix – Ox) = (2(m + n)u – (l – m – n)(v + w))/{(2(u + v + w)(l + m + n)}. We complete the parallelogram OhHPg, so that Pg has x-co-ordinate Ox + Hx – hx = l/(l + m + n). Its isogonal conjugate P has x-co-ordinate u(v + w)/l. This is a key step, for in a Hagge circle construction this figure is always a parallelogram. The plan is to show that the Hagge circle of all the triangles in the porism with respect to P coincide and have centre h, so that in particular, since the orthocentre of a triangle lies on all its Hagge circles, the locus of the orthocentre of all triangles in the porism is all or part of this Hagge circle. 3. Analysis of the special Hagge circle of triangle ABC generated by P The point Q, such that HhQ is a straight line and Hh = hQ, will lie on this Hagge circle and has co-ordinates Q(m + n – l, n + l – m, l + m – n)/(l + m + n).We show in due course that the equation of the Hagge circle is (l + m + n)(u(v + w)yz + v(w + u)zx + u(v + w)xy) – 2(x + y + z)(vwlx + wumy + uvnz) = 0.(3.1) It may be checked by substituting their co-ordinates in Equation (3.1) that H and Q do indeed lie on this circle. The points X, Y, Z which are the intersections of AH, BH, CH with the proposed Hagge circle are X((v + w)(m + n – l), 2vl, 2wl), Y(2um, (w + u)(n + l – m), 2wm), Z(2un, 2vn, (u + v)(l + m – n)). The equation of AP is vnz(w + u) = wmy(u + v) and this meets the circumcircle at the point D with co-ordinates D(– u(v + w), v(w + u)(m + n)/m, w(u + v)(m + n)/n). Likewise E and F have co-ordinates E(u(v + w)(n + l)/l, –v(w + u), w(u + v)(n + l)/n), F(u(v + w)(l + m)/l, v(w + u)(l + m)/m, – w(u + v)). Now the circle HBC has equation u(v + w)yz + v(w + u)zx + w(u + v)xy = 2vwx(x + y + z)
(3.2)
This meets the proposed Hagge circle at U with co-ordinates U(u(v + w), ((m + n)w + (n – m)u)v/m, ((m + n)v + (m – n)u)w/n). Similarly V and W have coordinates V((n + l)w +(n – l)v)u/l, v(w + u),((n + l)u +(l – n)v)w/n) W((l + m)v +(m – l)w)u/l, ((l + m)u + (l – m)w)v/m, w(u + v)).
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The co-ordinates of D and U are normalised to the same amount, so the displacement DU is proportional to ((v + w), – v, – w). The displacement BC is (0, – 1, 1), and the perpendicularity condition in areals is satisfied for these displacements. For those unfamiliar with this condition it is that (p, q, r) and (e, f, g) are perpendicular displacements if, and only if, u(v + w)(qg + rf) + v(w + u)(pg + re) + w(u + v)(pf + qe) = 0. See Bradley [4] p.180. Clearly the midpoint of DU lies on BC. It follows that D and U are reflections of one another in BC. Similar considerations hold for the pairs of points E, V and F, W relative to the sides CA, AB respectively. Thus the circle with equation (3) is the Hagge circle of P with respect to triangle ABC. As an aside it is interesting to consider the transformation, that we call the Hagge map. This takes (A, B, C) to (X, Y, Z). As it is known that this is an indirect similarity through the point P it is a matrix whose columns contain the normalised co-ordinates of X, Y, Z respectively. It has been checked by DERIVE that this matrix does indeed leave P invariant and maps triangle DEF on to triangle UVW. This provides a double check on the analysis of this section. 4. Analysis for an arbitrary triangle in the porism We now come to analyse an arbitrary triangle RST in the porism, where these points have parameters r, s, t respectively. The equation of ST is x/{u(v+w)} + sty/{v(w+u)} + (1 – s)(1 – t)z/{w(u+v)} = 0. (4.1) ST is a tangent to the inconic if, and only if, lu(v + w) + mv(w + u)/(st) + nw(u + v)/{(1 – s)(1 – t)}= 0, this being the condition for equal roots when Equation (4.1) is substituted in Equation (2.2). If we fix r then we know that s, t are the roots of the equation lu(v + w) + mv(w + u)/(rx) + nw(u + v)/{(1 – r)(1 – x)}= 0. Rearranging this gives (s + t) = (– r2lu(v+w) + r(lu(v + w) + mv(w + u) + nw(u + v)) – mv(w + u))/{r(1 – r)lu(v + w)} and st = – mv(w + u)/{rlu(v + w)}. These equations may be compactly re-expressed as rst = – mv(w + u)/{lu(v + w)} and (1 – r)(1 – s)(1 – t) = – nw(u + v)/{lu(v + w)}.
(4.2)
Essentially these equations say that triangle RST is in the porism, if, and only if, these two equations hold. Equation (4.1) may be written, using Equations (4.2) as lr(r – 1)x + m(1 – r)y + rnz = 0. (4.3) 4
A displacement (α, β, γ) along ST (that is, ST is proportional to (α, β, γ)) is therefore given by α = m(1 – r) – nr, β = nr – lr(r – 1), γ = lr(r – 1) – m(1 – r). (4.4) A rather tedious calculation shows that the line RP meets the circumcircle again at the point D' with parameter j = {m(l(r – 1) – n)}/{l(m(r – 1) + nr)}. (4.5) In terms of j the unnormalised co-ordinates of D' are (d, e, f) where d = – u(v + w)j(1 – j) , e = v(w + u)(1 – j), f = w(u + v)j.
(4.6)
We suppose that the reflection of D' in ST is U'(g, p, q) and we further suppose that U' is normalised with the same normalisation constant as D', so that g + p + q = d + e + f. (4.7) The midpoint ½(g + d, p + e, q + f) of D'U' lies on ST so we have from Equation (4.3) lr(g + d)(r – 1) + m(p + e)(1 – r) + nr(q + f) = 0.
(4.8)
The perpendicularity condition in this case is u(v + w)(β(q – f) + γ(p – e)) + v(w + u)(α(q – f) + γ(g – d)) + w(u + v)(α(p – e) + β(g – d)) = 0.
(4.9)
Values of α, β, γ in terms of l, m, n, r from Equations (4.4) may be substituted in Equation (4.9). Equations (4.5) and (4.6) may also be used in Equations (4.7) – (4.9), so that one is left with three equations for g, p, q in terms of l, m, n, r, u, v, w. Using DERIVE we find the (unnormalised) solution to be g = mnu(l2r(r – 1)(v + w) + l(v – w)(m(r – 1) + nr) + mn(v + w)), (4.10) p = –nlv(lr(m(r – 1)(u – w) – nr(u + w)) + m(m(r – 1)(u + w) + nr(w – u))), (4.11) q = lmw(l(r – 1)(m(r – 1)(u + v) + nr(v – u)) – n(m(r – 1)(u – v) – nr(u + v)))(4.12) It may now be determined by direct substitution that for any value of r the point U'(g, p, q) lies on the circle with Equation (3.1). By symmetry points V', W', similarly defined also lie on this circle. But the points U', V', W' define a Hagge circle, which contains H', the orthocentre of triangle RST. We have therefore proved that the locus of the orthocentres of the triangles in the porism is a circle or part of a circle, this circle being the Hagge circle of P with respect to each and every triangle in the porism. Consequently the porism is defined either by specifying the circumcircle, one triangle in the porism, say ABC, and the incentre I of an inconic. Or it is defined by specifying the point P instead of the point I. In the latter event the triangles in the porism may be 5
obtained as follows. Select a point H' on the Hagge circle of P with respect to triangle ABC. Obtain Pg', the point that is going to be the isogonal conjugate of P with respect to the triangle RST that has orthocentre H'. This is done by using the known fact that H'Pg'Oh is a parallelogram. The position of R may now be determined from the fact that angle ORP = angle H'RPg'. The positions of S and T then follow from the fact that ST is the perpendicular bisector of H. Reference 1. E. A. Maxwell, The methods of plane projective geometry based on the use of general homogeneous coordinates. Cambridge University Press, 1946. 2. C.J. Bradley & G.C.Smith, Stationary Porisms, Accepted for publication.
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Article 28 Some Circles in a Cyclic Quadrilateral Christopher J Bradley H
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1. Introduction In the figure ABCD is a cyclic quadrilateral, Points T, U, V, W are the midpoints of AB, BC. CD, DA respectively and O is the centre, X is the midpoint of AC and Z is the midpoint of BD. The diagonals AC and BD meet at Y. Points P, Q, R, S are the midpoints of OA, OB, OC, OD 1
respectively, so that PQRS is a cyclic quadrilateral similar to ABCD and half the size. The following results now hold, several of which are immediate and others more interesting. (i)
Circles ATW, BUT, CVU, DWV all pass through O and touch the circle ABCD at its vertices A, B, C, D respectively; (ii) Circles ATW, CVU pass through X and circles BUT, DWV pass through Z; (iii) Points O, X, Y, Z are concyclic; (iv) Circles AYZ and COZ meet at Z and a point H lying on the circumcircle; (v) Circles CYZ and AOZ meet at Z and a point F lying on the circumcircle; (vi) Circles BXY and DOX meet at X and a point E lying on the circumcircle; (vii) Circles DXY and BOX meet at X and a point G lying on the circumcircle. (viii) Angle OXY = angle OZY = 90o. In the following sections we prove these results using Cartesian co-ordinates. 2. Result (i) Take ABCD as the unit circle with O the origin. Let A have co-ordinates (2a/(1 + a2), (1 – a2)/(1 + a2) and let B. C, D have similar co-ordinates with parameters b, c, d respectively. The coordinates of T, the midpoint of AB, are therefore {(1 + ab)/((1 + a2)(1 + b2))}(a + b, 1 – ab). Similar expressions hold for the co-ordinates of U, V, W by making appropriate changes of parameters. The equation of the circle ATW may now be calculated and is (1 + a2)(x2 + y2) – 2ax – (1 – a2)y = 0. (2.1) Note that this circle passes through the point O and has centre P, the midpoint of AO. Similarly the circles BUT, CVU, DWV all pass through O and have centres at Q, R, S respectively. An alternative argument for this result is that the four circles are the Miquel circles for the four vertices A, B, C, D and must have a common point. And furthermore the quadrangle of centres must be similar to the quadrangle ABCD, and since it derives from the midpoints of chords the similarity is simply a reduction by a factor of 2 through the Miquel point O. It is also easy to check that circle ATW touches circle ABCD at A. 3. Result (ii) The circle CVU has an equation similar to Equation (2.1) with c replacing a. It follows trivially that the circles ATW, CVU meet at the midpoint X of AC as well as at O. Similarly circles BUT and DWV meet at the midpoint Z of BC as well as at O. The co-ordinates of X and Z are similar to those of T but with appropriate changes of parameters. 2
4. Result (iii) The point Y is the intersection of the diagonals AC and BD. The line AC has equation (a + c)x + (1 – ac)y = 1 + ac.
(4.1)
The line BD is similar with b, d replacing a, c. These two lines meet at Y with co-ordinates (x, y), where x = 2(ac – bd)/(abc + acd – abd – bcd + a – b + c – d), y = (abc + acd – abd – bcd – a + b – c + d)/(abc + acd – abd – bcd + a – b + c – d).
(4.2) (4.3)
Having obtained the co-ordinates of X, Y, Z we may now obtain the equation of circle XYZ and this is found to be (abc + acd – abd – bcd + a – b + c – d)(x2 + y2) + 2(bd – ac)x + (abc + acd – abd – bcd – a + b – c + d)y = 0. (4.4) Clearly this circle passes through O. 5. Result (iv) The circle AZY has equation (abc + acd – abd – bcd + a – b + c – d)(x2 + y2) + (ab + ad – 2ac + 2bd – bc – cd)x + (abc – 2abd + acd + b – 2c + d)y – (a – c)(1 + bd) = 0. (5.1) The circle COZ has equation (bc2 – 2bcd + c2d – b + 2c – d)(x2 + y2) + 2x(bd – c2) + (bc2 – 2bcd + c2d + b – 2c + d) = 0. (5.2) These circles meet at Z and the point H with co-ordinates (x, y), where x = – 2{(b – 2c + d)(bc + cd – 2bd)}/{b2(c2 – 4cd + 4d2 + 1) + 2b(c2d – 2c(1 + d2) + d) + c2(4 + d2) – 4cd + d2}, (5.3) y = – {(b(c – 2d + 1) + c(d – 2) + d)(b(c – 2d – 1) + c(d + 2) – d)}/ )}/{b2(c2 – 4cd + 4d2 + 1) + 2b(c2d – 2c(1 + d2) + d) + c2(4 + d2) – 4cd + d2}. (5.4) It may now be verified that (x, y) given by Equations (5.3) and (5.4) satisfy x 2= y2 = 1 so that circles AZY and COZ intersect at Z and a point H on the circumcircle of ABCD. Results (v), (vi), (vii) now follow by symmetry. 6. Result (viii) 3
It is straightforward to show the co-ordinates of the centre K of circle OXYZ are half those of point Y, which proves that O, K, Y are collinear and hence angles OZY and OXY are right angles. 7. The cyclic pentagon In Figure 2 we show a cyclic pentagon, in which the same construction is carried out for each of the five component cyclic quadrilaterals. There are five diagonals whose midpoints are labelled X1 › X5 and whose intersections (other than A, B, C, D, E) are labelled Y1› Y5. There are thus 5 Miquel circles touching the circumcircle at its vertices and 5 more circles such as OX1Y2X2. There would also be 40 circles intersecting in pairs on the circumcircle. For clarity we show only one of these pairs in Figure 2, AY2X2^COX2 = H2.
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ARTICLE 29 The Miquel Circles for a Quadrilateral Christopher J Bradley
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1. Introduction Let ABCD be a quadrilateral in which no pair of opposite sides are parallel, and let AD and BC meet at F and BA and CD meet at G. In triangle BFG, points V, W, J are taken on BF, BG, FG respectively. Now let M be the Miquel point for this configuration, so that circles JFV, JGW, BVW meet at M. Suppose now circle JFV meets ADF at T and circle JGW meets GDC at U. Now consider triangle FAB with T on FA, V on FB and W on AB. Since circles BVW and FJV are known to meet at M, it follows that M is the Miquel point for this configuration 1
and it follows circle ATW also passes through M. Similarly by considering triangle GBC, it follows that circle CVU passes through M. The following results now hold: (i) Circle DTU passes through M; (ii) Circles ATW and CVU meet at M and a point L lying on AC; (iii)Circles BWV and DTU meet at M and a point N lying on BD; (iv) If E is the intersection of the diagonals AC and BD, then L, E, M, N are concyclic. Whereas it is true that results (i), (ii), (iii) can be proved by synthetic arguments, there does not seem to be a pure argument of (iv). And unfortunately the analytic proof of (iv) requires an analysis of everything else first and this means a technically difficult proof. We use areal co-ordinates with B(1, 0, 0), F(0, 1, 0), G(0, 0, 1) so that BFG is triangle of reference. 2. Points J, C, D, A, M We take C to be the point C(m, 1 – m, 0), A to be the point A(1 – n, 0, n) and J to be the point J(0, k, 1 – k). The equation of CG is (1 – m)x = my, (2.1) and the equation of the line AF is (1 – n)z = nx.
(2.2)
CG and AF meet at D(m(1 – n), (1 – m)(1 – n), mn). Rather than designate co-ordinates for points such as T, U, V, W, it is easier to let M have co-ordinates (r, s, t) and to find T, U, V, W in terms of r, s, t and other variables. Now since we are not dealing with normalised co-ordinates or the ratio of lengths we can choose r + s + t to be whatever we like, without altering the position of M. We therefore choose r + s + t = 1/k, so that J can be made to alter its position without M moving. This simplification still allows M and J to have independent positions. 3. Circles FJM, GJM and points T, U, V, W In this section we give no working, but simply display the results as output by the algebraic computer package DERIVE that we used, the aim being to prevent errors in typing. The equation of the circle FJM is – x2 (a2t(s – 1) + r(b2t + c2s)) – x(y(a2t(s – 1) + r(b2t – c2(r + t))) + z(a2(r + st – t) – r(b2(r + s) – c2s))) + a2rz(y(r + s + t – 1) – z) = 0. (3.1) The circle GJM has equation x2(a2s(r + s – 1) – r(b2t + c2s)) + x(z(a2s(r + s – 1) + r(b2(r + s) – c2s)) – y(a2(r2 + r(t – 1) – s2 + s) + r(b2t – c2(r + t)))) – a2ry(y(r + s + t – 1) – z) = 0. (3.2) The co-ordinates of T are (x, y, z), where x = (n – 1)(a2(n(r2 + r(s + t – 1) + st – t) –st + t) + r(n – 1)(b2t – c2( r + t))), y = r(1 – n)(b2(n(r + s + t) – t) – c2s) – a2(n(r – st + t) + st – t), z = – n(a2(n(r2 + r(s + t – 1) + st – t) – st + t) + r(n – 1)(b2t – c2(r + t))). 2
(3.3)
The co-ordinates of V are (x, y, z), where x = a2t(1 – s) – r(b2t – c2(r + t)), y = a2t(s – 1) + r(b2t + c2s), z = 0. The co-ordinates of W are (x, y, z), where x = a2s(r + s – 1) + r(b2(r + s) – c2s), y = 0, z = r(b2t + c2s) – a2s(r + s – 1).
(3.4)
(3.5)
The co-ordinates of U are (x, y, z), where x = – m(a2(m(r + s)(s – 1) + r) + mr(b2(r + s) – c2s)), y = (m – 1)(a2(m(r + s)(s – 1) + r) + mr(b2(r + s) – c2s)), (3.6) z = a2(m(r2 + r(2s + t – 1) + s(s – 1)) – r(r + s + t – 1)) – mr(b2t + c2(m(r + s + t) – r – t)). 4. The circles through A, B, C, D In Section 3 we obtained points T, U, V, W lying on the lines AD, DC, CB, BA respectively. These points cannot be independent of one another, for if one places points on the sides at random then the four circles ATW, BWV, CVU, DUT do not pass through a common point. In other words the circles FJM, GJM form a recipe for obtaining four points T, U, V, W having the property that circles ATW, BWV, CVU, DUT share a common point M. In this way, for any quadrilateral without a pair of parallel sides, it is shown how to construct a Miquel point and the four associated circles. In this section we find the equations of the four circles and check that they do share the common point M with circles FJM and GJM. The equation of circle BWV is x(y(a2t(s – 1) + r(b2t + c2s)) + z(r(b2t + c2s) – a2s(r + s – 1))) + y2(a2t(s – 1) + r (b2t – c2(r + t))) + yz(a2(r2 + rt + (1 – s)(s – t)) – r(b2(r + s – t) + c2(r – s + t))) – z2(a2s(r + s – 1) + r(b2(r + s) – c2s)) = 0. (4.1) The equation of circle DUT is x2(a2(m(n(rs + (s – 1)(s – t)) + t(s – 1)) + t(n – 1)(s – 1)) – r(b2t + c2s)(m(2n – 1) – n + 1)) + x(z(a2(m(n(r(2s – 1) + (s – 1)(2s – t)) + (1 – s)(r + s – t)) + (n – 1)(r + t(s – 1))) + r(b2(m(n(2r +2s – t) – r – s + t) – (n – 1)(r + s)) – c2s(m(3n – 2) – n + 1))) – y(a2(m(n(r2 + r(t – 1) + (1 – s)(s – 2t)) – 2t(s – 1)) + t(1 – s)(n – 1)) + r(b2t(m(3n – 2) – n + 1) + c2((n – 1)(r + t) – m(n(2r – s + 2t) – r + s – t))))) – my2(a2(n(r2 + r(s + t – 1) + t(s – 1)) – t(s – 1)) + r(n – 1)(b2t – c2(r + t))) + yz(a2(m(n(rs + (s – 1)(s – t)) + (1 – s)(r + s – t)) – r(n – 1)(r + s + t – 1)) + mr(n – 1)(b2(r + s – t) + c2(r – s + t))) +z2(n – 1)(a2(m(r + s)(s – 1) + r) + mr(b2(r + s) – c2s)) = 0. (4.2) The equation of circle ATW is 3
nx2 (a2s(r + s – 1) – r(b2 t + c2 s)) + x(z(a2s(r + s – 1)(2n – 1) + r(b2(n(r + s – t) + t) + c2s(1 – 2n))) – y(a2(n(r 2 + r(t – 1) + (1 – s)(s – t)) – t(s – 1)) + r(b2 t(2n – 1) – c2 (n(r – s + t) + s)))) – y2(a2 (n(r2 + r(s + t – 1) + t(s – 1)) – t(s – 1)) + r(n – 1)(b2 t – c2 (r + t))) – yz(a2(n(r2 + r(t – 1) + (1 – s)(s – t)) – r2 – rt + (s – 1)(s – t)) + r(1 – n)(b2 (r + s – t) + c2 (r – s + t))) + z2 (n – 1)(a2 s(r + s – 1) + r(b2 (r + s) – c2 s)) = 0. (4.3) The equation of circle CUV is x2 (m – 1)(a2 t(s – 1) + r(b2 t + c2s)) + x(y(a2 t(s – 1)(2m – 1) + r(b2t(2m – 1) – c2(m(r – s + t) – r – t))) – z(a2 (m(r + s – t)(s – 1) + r + t(s – 1)) + r(b2 (m(r + s – t) – r – s) + c2 s(1 – 2m)))) + my2(a2t(s – 1) + r(b2t – c2(r + t))) – yz(a2(m(r + s – t)(s – 1) – r(r + s + t –1)) + mr(b2(r + s – t) + c2(r – s + t))) – z2(a2(m(r + s)(s – 1) + r) + mr(b2(r + s) – c2s)) = 0. (4.4) It may now be checked that all four circles pass through M(r, s, t). 5. The points N, L, E and the circle LENM The point N lies on BD and is the intersection of circles DUT and BVW. Its co-ordinates are (x, y, z), where x = a2(m2 (n2 (r2 + r(t – s) + 2(1 – s)(s – t)) – n(r2 + rt + (1 – s)(s – 3t)) + t(s – 1)) + m(1 – n)(n(r2 + rt + (1– s)(s – 3t)) – 2t(s – 1)) + t(n – 1)2 (s – 1)) – r(b2(m(n(r + s – t) + t) + t(n – 1)) + c2(m(n(r – s + t) – r – t) – (n – 1)(r + t)))(m(2n – 1) – n + 1), y = (m – 1)(n – 1)(a2(m(n(rs + (s – 1)(s – t)) + t(s – 1)) + t(n – 1)(s – 1)) – r(b2t + c2s)(m(2n – 1) – n + 1)), (5.1) z = mn(a2(m(n(rs + (s – 1)(s – t)) + t(s – 1)) + t(n – 1)(s – 1)) – r(b2t + c2s)(m(2n – 1) – n + 1)). The point L lies on AC and is the intersection of circles ATW and CUV. Its co-ordinates are (x, y, z), where x = a2(m2(s – 1)(n(r + s + t) – t) + m(n2(r2 + r(2s + t – 1) + s2 + s(t – 1) – t) – n(r2 + r(2s + t – 2) + s2 + s(3t – 1) – 3t) + 2t(s – 1)) – n2(r2 + r(s + t – 1) + t(s – 1)) + n(r2 + r(s + t – 1) + 2t(s – 1)) – t(s – 1)) + r(m + n – 1)(b2(m(n(r + s + t) – t) – t(n – 1)) + c2((n – 1)(r + t) – m(n(r + s + t) – r – t))), y = (1 – m)(a2(m(s – 1)(n(r + s + t) – t) + n(r – t(s – 1)) + t(s – 1)) + r(m + n – 1)(b2(n(r + s + t) – t) – c2s)), (5.2) z = n(r(m + n – 1)(b2t + c2(m(r + s + t) – r – t)) – a2(m(n(r2 + r(2s + t – 1) + (s – 1)(s + t)) – t(s – 1)) – n(r2 + r(s + t – 1) + t(s – 1)) + t(s – 1))). The point E is the intersection of the diagonals AC and BD and has co-ordinates (x, y, z), where x = 2n(1 – n), y = (1 – m)(1 – n), (5.3) z = mn. 4
It may now be verified that points L, E, N, M all lie on the circle with equation x2(a2(m(n(rs + s2 – s(t + 1) + t) + t(s – 1)) + t(n – 1)(s – 1)) – r(b2t + c2s)(m(2n – 1) – n + 1)) + x(z(a2(m(n(r(3s – 1) + 3s2 – s(t + 3) + t) + r(1 – 2s) + (1 – s)(2s – t)) + (n – 1)(r + t(s – 1))) + r(b2(m(2n(r + s – t) – r – s + 2t) – (n – 1)(r + s)) – c2s(m(4n – 3) – n + 1))) – y(a2(m(n(r2 + r(t – 1) – s2 + s(3t + 1) – 3t) – 3t(s – 1)) + t(1 – s)(n – 1)) + r(b2t(m(4n – 3) – n + 1) + c2((n – 1)(r + t) – m(2n(r – s + t) – r + 2s – t))))) – my2(a2(n(r2 + r(s + t – 1) + 2t(s – 1)) – 2t(s – 1)) + 2r(n – 1)(b2 t – c2(r + t))) – yz(a2(m(n(r2 + r(t – s) + 2(1 – s)(s – t)) – r2 + r(s – t – 1) +2(s – 1)(s – t)) + r(n – 1)(r + s + t – 1)) + 2mr(1 – n)(b2(r + s – t) + c2(r – s + t))) + z2(n – 1)(a2(m(r(2s – 1) + 2s(s – 1)) + r) + 2mr(b2 (r + s) – c2 s)) = 0. (5.4) 6. Other quadrilaterals The same construction of the Miquel point is valid with minor simplifications for quadrilaterals that have one or two pairs of opposite sides parallel. For a trapezium with AD parallel to BC, F recedes to infinity and the circle FJM becomes a line. For a parallelogram both F and G recede to infinity and both circles FJM and GJM become lines. The result is that TMV and WMU become straight lines.
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Article 30 More on Circular Perspective Christopher J Bradley
W
F
A Y E C
B
U D Z
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Figure
1. Introduction
1
When the vertices of two triangles ABC and UVW (with no vertices of ABC on the sides of UVW and vice versa) are such that circles AVW, BWU, CUV share a common point, then circles UBC, VCA, WAB also share a common point and the circles are said to be in circular perspective. If furthermore circles AWU, BUV, CVW share a common point then circles AUV, BVW, CWU also share a common point; and consequently circles UCA, VAB, WBC share a common point and circles UAB, VBC, WCA also share a common point. Then triangles ABC and UVW are said to be in triple circular perspective. See Article 19. A singular case occurs when the vertices U, V, W lie on the sides BC, CA, AB respectively and all of the above holds, except that ‘circles’ UBC, VCA, WAB are actually straight lines (to be thought of as having the complex point ∞ in common). An even more singular case occurs when UVW is a transversal of triangle ABC and in this article we provide an instance of this singular case when triple circular perspective holds. The case we investigate is when U is the intersection of the tangent at A to the circumcircle of triangle ABC with the side BC, with V and W similarly defined on the tangents at B and C and on the sides CA and AB respectively. See the figure above. The surprise is that the five finite points of intersection lie, three of them on the circumcircle of ABC (the points D, E, F in the figure) and two on the line UVW (the points Y, Z in the figure). That one of the three points should be on the circumcircle is because the Miquel point of circles AVW, BWU, CUV is bound to do so, but the other two are not Miquel points and hence the surprise. Analytic proofs are provided in the following sections, using areal co-ordinates with ABC as triangle of reference. 2. The tangents and the points U, V, W The equation of the tangent at A is c2y + b2z = 0 and this meets BC, x = 0, at the point U with coordinates (0, b2, – c2). Similarly V has co-ordinates (– a2, 0, c2) and W has co-ordinates (a2, – b2, 0). U is, of course, the harmonic conjugate of the point L on BC, with respect to B and C, where the line AK meets BC, K being the symmedian point of ABC. The equation of the line UVW is easily seen to be x/a2 + y/b2 + z/c2 = 0 and hence UVW is a transversal of triangle ABC. 3. Circles ABV, BCW, CAU and the point Y The equation of the circle ABV is 2
(2.1)
a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(3.1)
where u, v, w are to be chosen so that A, B and V lie on the curve. A short analysis gives u = 0, v = 0 and w = a2b2/(c2 – a2). Its intersection with the line having Equation (2.1) is a point with coordinates (x, y, z), where x = a2(a2 – b2), y = b2(b2 – c2), (3.2) 2 2 2 z = c (c – a ). Since these co-ordinates are invariant under the cyclic change of x, y, z and a, b, c, this point also lies on circles BCW and CAU and is therefore the point Y in the figure. 4. Circles ABU, BCV, CAW and the point Z A similar analysis to that of Section 3 gives u = 0, v = 0, w = a2b2/(c2 – b2) for the circle ABU. The intersection of circle ABU and the line with Equation (2.1) is a point with co-ordinates (x, y, z), where x = a2(a2 – c2), y = b2(b2 – a2), (4.1) 2 2 2 z = c (c – b ). Again since these co-ordinates are invariant under the cyclic change of x, y, z and a, b, c, this point also lies on circles BCV and CAW and is therefore the point Z in the figure. 5. Circles AUV, BVW, CWU and the point F The circle AUV has Equation (3 .1) with u = 0, v = a2c2(a2 – b2)/{(a2 – c2)(b2 – c2)}, w = a2b2/(c2 – a2) and this circle cuts the circumcircle with equation a2yz + b2zx + c2xy = 0, (5.1) at A and the point with co-ordinates (x, y, z), where x = a2(a2 – b2)(b2 – c2), y = b2(b2 – c2)(c2 – a2), z = c2(c2 – a2)(a2 – b2).
(5.2)
Again since these co-ordinates are invariant under the cyclic change of x, y, z and a, b, c, this point also lies on circles BVW and CWU and is therefore the point F in the figure. 6. The points D and E
3
Further analysis shows that circles AVW, BWU, CUV meet at the point D on the circumcircle with co-ordinates (x, y, z), where x = a2(a2 – b2)(c2 – a2), y = b2(b2 – c2)(a2 – b2), (6.1) 2 2 2 2 2 z = c (c – a )(b – c ). Likewise, circles AWU, BUV, CVW meet at the point E on the circumcircle with co-ordinates (x, y, z), where x = a2(b2 – c2)(c2 – a2), y = b2(c2 – a2)(a2 – b2), (6.2) 2 2 2 2 2 z = c (a – b )(b – c ). In conclusion it should be mentioned that not all transversals show triple circular perspective. Acknowledgement My thanks to David Monk who has communicated with me and informed me of a second case of triple circular perspective with a transversal when U is the intersection of BC with the bisector of the exterior angle at A, with V and W similarly defined.
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Article 31 More cases of Circular Perspective Christopher J Bradley
A M
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bH
aH
'
H cH
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Figure 1
Commentary This article should be read in conjunction with Article 19. There it is proved, but not explicitly, that circles ABΩ, CAΩ', BCH are concurrent at aH, circles BCΩ, ABΩ', CAH are concurrent at bH and circles CAΩ, BCΩ', ABH are concurrent at cH. It follows that triangles ABC and ΩHΩ' are in triple circular perspective. This means, additionally, that circles AΩH, BHΩ', CΩ'Ω are concurrent at a point L, circles AHΩ', BΩ'Ω, CΩH are concurrent at a point M and circles AΩ'Ω, BΩH, CHΩ' are concurrent at a point N. See the figure above.
2
ABC and OΩΩ' are in Triple Circular Perspective
M A
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O Q
'
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B C
L
Figure
3
1. Introduction We now establish a very interesting result, namely that triangles ABC and ΩΩ'O are in triple circular perspective, but not only that, it is also the case that the six points of concurrence lie three on the seven-point circle (which is the circumcircle of triangle ΩΩ'O) and three on the circumcircle of ABC. One is led to wonder if this configuration is essentially unique for the six points of concurrence to lie three by three on the two circumcircles. In the following analysis we use areal co-ordinates with triangle ABC as triangle of reference. 2. The circles ABO, BCΩ, CAΩ' and the points P, Q, R The general equation of a circle is a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(2.1)
where u, v, w are constants to be determined. One then inserts the co-ordinates of the three points on the given circle to provide equations for u, v, w. When this is done for circle ABO we find that u = 0, v = 0 and w = – a2b2/(a2 + b2 – c2) and hence its equation is – a2b2z2 – a2(c2 – a2)yz + b2(b2 – c2)zx + c2(a2 + b2 – c2)xy = 0. (2.2) Using the same method we find the equation of circle BCΩ is – b2x2 + a2yz + (c2 – b2)xy = 0.
(2.3)
These circles meet at a point R with co-ordinates (a2, b2, a2 + b2 – c2). See the figure above. Now the equation of CAΩ' is – a2y2 + b2zx + (c2 – a2)xy = 0.
(2.4)
It may now be verified that R lies also on this circle. It follows now by cyclic change of letters that circles BCO, CAΩ, ABΩ', all pass through the point P with co-ordinates (b2 + c2 – a2, b2, c2) and that circles CAO, ABΩ, BCΩ' all pass through the point Q with co-ordinates (a2, c2 + a2 – b2, c2). The equation of the seven-point circle is b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0. It may now be checked that P, Q, R all lie on the seven-point circle. 3. Circles OΩC, ΩΩ'A, Ω'OB and the points L, M, N
4
(2.5)
Using the same method as in Section 2 we find the equation of circle OΩC to be b2c2(b2 – c2)x2 – c2a2(a2 – b2)y2 + a2(a4 – a2(b2 – c2) + c4)yz + a2b2(a2 – b2)zx + c2(b2 – c2)2xy = 0. (3.1) Also the equation of circle ΩΩ'A is – c2a2(a2 – b2)y2 + a2b2(c2 – a2)z2 – a2(a2 – b2)(c2 – a2)yz + b2c2(a2 – b2)zx – b2c2(c2 – a2)xy = 0. (3.2) 2 2 2 2 2 2 2 2 2 These two circles meet at the point L with co-ordinates (a (a – b )(c – a ), b (c – a )(b – c2), c2(b2 – c2)(a2 – b2)). Now the equation of the circle Ω'OB is b2c2(b2 – c2)x2 – a2b2(c2 – a2)z2 – a2(a4 – a2(b2 + c2) + b4)yz – b2(b2 – c2)2zx + c2a2(c2 – a2)xy = 0. It may now be checked that L lies also on this circle.
(3.3)
The equations of the circles OΩA, ΩΩ'B, Ω'OC may now be written down with cyclic change of letters and verified that they all pass through the point M with co-ordinates (a2(c2 – a2)(b2 – c2), b2(b2 – c2)(a2 – b2), c2(a2 – b2)(c2 – a2)). Similarly the circles OΩB, ΩΩ'C, Ω'OA all pass through the point N with co-ordinates (a2(b2 – c2)(a2 – b2), b2(a2 – b2)(c2 – a2), c2(c2 – a2)(b2 – c2)). Finally it may be verified that L, M, N all lie on the circumcircle of ABC, whose equation is a2yz + b2zx + c2xy = 0. (3.4)
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ARTICLE 32 The GH Disc and another case of Triple Circular Perspective Christopher J Bradley
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1. Introduction
1
C
This article should be read after Articles 19, 30, 31, in which circular perspectivity is defined and where particular emphasis is given to triple circular perspectivity and certain examples of this property. In this article we take three points U, V, W on the orthocentroidal circle S and show that triangle UVW is in triple circular perspective with ABC. Circles BUC, CVA, AWB meet at the orthocentre H, circles BVC, CWA, AUB meet at the first Brocard point Ω and circles BWC, CUA, AVB meet at the second Brocard point Ω'. Also circles AVW, BWU, CUV meet at a point L on the circumcircle of ABC, its counterpart H lying on the circumcircle of UVW (the orthocentroidal circle). In the above the point U may be defined as follows. Draw AH to meet S again at X, then draw XK to meet S again at U. Here K is the symmedian point. V, W are similarly defined. Alternatively U, V, W are the points on the Hagge circle of K (which coincides with S) such that U, V, W are the reflections of D, E, F in BC, CA, AB respectively, where D, E, F are the intersections of AK, BK, CK with the circumcircle of ABC. 2. The orthocentroidal circle and the points U, V, W The equation of the orthocentroidal circle S, see Bradley and Smith [1], is (b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2 – a2yz – b2zx – c2xy = 0.
(2.1)
The equation of the line AH is (c2 + a2 – b2)y = (a2 + b2 – c2)z.
(2.2)
This meets S at the point X with co-ordinates (a2, a2 + b2 – c2, c2 + a2 – b2). The symmedian point K has co-ordinates (a2, b2, c2) and so XK has equation (b2 + c2 – a2)(b2 – c2)x + a2(a2 – b2)y + a2(c2 – a2)z = 0. (2.3) This meets the circle S again at U with co-ordinates (a2, b2 + c2 – a2, b2 + c2 – a2). Similarly V has co-ordinates (c2 + a2 – b2, b2, c2 + a2 – b2) and W has the co-ordinates (a2 + b2 – c2, a2 + b2 – c2, c2). (It will be seen these points are those called aH, bH, cH in Article 19.) 3. Circles BUC, CVA, AWB all pass through H The equation of the circle BUC is (b2 + c2 – a2)x2 – a2yz + (c2 – a2)zx – (a2 – b2)xy = 0.
(3.1)
It may now be checked that this circle passes through H and by cyclic change of x, y, z and a, b, c it follows that circles CVA and AWB also pass through H. 4. Circles BVC, CWA, AUB all pass through Ω The equation of circle BVC is 2
b2x2 – a2yz + (b2 – c2)xy = 0
(4.1)
and the equation of CWA is c2y2 + (c2 – a2)yz – b2zx = 0.
(4.2)
These circles meet at the point Ω(1/b2, 1/c2, 1/a2). Similarly circles BWC. CUA, AVB meet at the point Ω'(1/c2, 1/a2, 1/b2). 5. Circles BWU, CUV, AVW meet on the circumcircle of ABC The equation of the circle BWU turns out to be c2(b2 – c2)(b2 + c2 – a2)x2 + a2(a2 – b2)(a2 + b2 – c2)z2 + c2a2(b2 – c2)yz – (a6 – a4(b2 + c2) – a2(b4 – 3b2c2 + c4) + (b2 + c2)(b2 – c2)2)zx + c2a2(a2 – b2)xy = 0.
(5.1)
The equations of circles CUV and AVW may now be written down by cyclic change of x, y, z and a, b, c. These circles all meet at L on the circumcircle of ABC with co-ordinates (x, y, z), where x = a2/((b2 – c2)(b2 + c2 – a2)), y = b2/((c2 – a2)(c2 + a2 – b2)), z = c2/((a2 – b2)(a2 + b2 – c2)).
(5.2)
Circles BUV, CVW, AWU meet at M and circles BVW, AWU, AUV meet at N, see the figure. Reference 1. C.J.Bradley and G.C.Smith, The locations of triangle centres, Forum. Geom., 6 (2006) 57-70. Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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ARTICLE 33 29 Circles Christopher J Bradley
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B Q M
Figure
1
1. Introduction This article should be read in conjunction with Article 19, and 30 – 32. It involves two cases of triple circular perspective. The first is between triangle ABC and triangle ΩΩ'O, where O is the circumcentre of ABC and Ω, Ω' are respectively the first and second Brocard points of ABC. The resulting six points of concurrence are labelled L, M, N, U, V, W. The second is between ABC and triangle UVW and the sixpoints of concurrence are Ω, Ω', O, P, Q, R. Interestingly U, V, W lie on the 7-point circle defined by O,Ω and Ω' and all of L, M, N, P, Q, R lie on the circumcircle of ABC. It is also the case that U, V, W lie on BK, CK, AK respectively, where K is the symmedian point of ABC, which lies itself on the 7-point circle diametrically opposite O. This interaction between the circumcircle and the 7-point circle is especially interesting. See the figure above. Note also that U, V, W do not coincide with any of the usual 7 points on the 7-point circle and are therefore additional points of some significance. The analysis of the configuration just described is carried out in subsequent sections using areal co-ordinates with triangle ABC as triangle of reference. 2. The points L, M, N The equation of any circle in areal co-ordinates is a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
To obtain the equation of the circle CΩΩ' one puts into Equation (2.1) the co-ordinates of C, Ω, and Ω' in turn, solve for u, v, w and then re-inset their values back into the equation, yielding the result. It turns out to be b2c2(c2 – a2)x2 – c2a2(b2 – c2)y2 – a2b2(c2 – a2)yz + a2b2(b2 – c2)zx + c2(c2 – a2)(b2 – c2)xy = 0. (2.2) This circle meets the circumcircle of ABC again at the point L with co-ordinates (x, y, z), where x = a2(a2 – b2)(b2 – c2), y = b2(a2 – b2)(c2 – a2), (2.3) z = c2(c2 – a2)(b2 – c2). The equation of circle AΩ'O is c2a2(b2 – c2)y2 – a2b2(a2 – b2)z2 + a2(a2 – b2)2yz – b2c2(b2 – c2)zx + c2(a4 – a2c2 – c2(b2 – c2))xy = 0.
(2.4)
It may now be checked from Equations (2.3) and (2.4) that L also lies on this circle. Symmetry considerations now imply that L also lies on circle BOΩ. We next determine the co-ordinates of the point M. The equation of circle AΩΩ' is c2a2(a2 – b2)y2 – a2b2(c2 – a2)z2 + a2(a2 – b2)(c2 – a2)yz – b2c2(c2 – a2)zx + b2c2(c2 – a2)xy = 0.
(2.5)
This circle meets the circumcircle again at the point M with co-ordinates (x, y, z), where x = a2(a2 – b2)(c2 – a2), y = b2(b2 – c2)(c2 – a2), (2.6) 2
z = c2(a2 – b2)(b2 – c2). The equation of circle COΩ is b2c2(b2 – c2)x2 – c2a2(a2 – b2)y2 + a2(a4 – a2(b2 + c2) + c4)yz + a2b2(a2 – b2)zx + c2(b2 – c2)2xy = 0.
(2.7)
It may be checked from Equations (2.6) and (2.7) that M also lies on this circle and similarly on circle BΩ'O. We next determine the co-ordinates of the point N. The equation of circle BΩΩ' is b2c2(a2 – b2)x2 – a2b2(b2 – c2)z2 – c2a2(a2 – b2)yz – b2(b2 – c2)(a2 – b2)zx + c2a2(b2 – c2)xy = 0. This circle meets the circumcircle at the point N with co-ordinates (x, y, z), where x = a2(b2 – c2)(c2 – a2), y = b2(a2 – b2)(b2 – c2), z = c2(a2 – b2)(c2 – a2). The equation of circle AOΩ is c2a2(c2 – a2)y2 – a2b2(b2 – c2)z2 + a2(c2 – a2)2yz + b2(a4 – a2b2 + b2(b2 – c2))zx + b2c2(b2 – c2)xy = 0.
(2.8)
(2.9)
(2.10)
It may be checked from Equations (2.9) and (2.10) that N lies on this circle too and by symmetry considerations also on circle CΩ'O. 3. The points U, V, W We first consider circle AΩB whose equation is found to be a2z2 + (a2 – b2)zx – c2xy = 0.
(3.1)
Now the equation of the 7-point circle is b2c2x2 + c2a2y2 + a2b2c2 – a4yz – b4zx – c4xy = 0,
(3.2)
Circle AΩB meets the 7-point circle at the point U with co-ordinates (a2, c2 + a2 – b2, c2). Circle AOC is found to have equation c2a2y2 – a2(a2 – b2)yz + b2(c2 + a2 – b2)zx – c2(b2 – c2)xy = 0.
(3.3)
It may now be checked that this circle passes through the point U, and similarly circle BΩ'C passes through U. The equation of BΩC is b2x2 – a2yz +(b2 – c2)xy = 0.
(3.4)
From Equations (3.2) and (3.4) this meets the 7-point circle at the point V with coordinates (a2, b2, a2 + b2 – c2). 3
The equation of circle AOB is a2b2z2 + a2(c2 – a2)yz – b2(b2 – c2)zx – c2(a2 + b2 – c2)xy = 0. (3.5) It may now be checked that this circle passes through V and similarly circle CΩ'A passes through V. Finally the point W with co-ordinates (b2 + c2 – a2, b2, c2) lies on the 7-point circle and circle CΩA with equation c2y2 +(c2 – a2)yz – b2zx = 0. (3.6) It may now be confirmed that W also lies on circles BOC and AΩ'B. 4. The points P, Q, R The equation of circle AVW is c2(b2 – c2)y2 – b2(c2 – a2)z2 – (c2 – a2)2yz – b2(b2 – c2)zx + c2(c2 – a2)xy = 0. This meets the circumcircle again at the point P with co-ordinates (x, y, z), where x = a2(b2 – c2)(c2 – a2), y = b2(a2 – b2)(c2 – a2), z = c2(a2 – b2)(b2 – c2). The equation of circle CUV is b2(a2 – b2)x2 – a2(b2 – c2)y2 – a2(a2 – b2)yz + b2(b2 – c2)zx – (b2 – c2)2xy = 0.
(4.1)
(4.2)
(4.3)
It may now be checked that P lies on this circle and similarly also on circle BWU. The circle AUV has equation c2a2(c2 – a2)y2 – a2b2(a2 – b2)z2 – a2(a2 – b2)(c2 – a2)yz + b2(a2b2 – b4 + b2c2 – c4)zx + c2(a2c2 – b4 + b2c2 – c4)xy = 0. (4.4) This circle meets the circumcircle again at the point Q with co-ordinates (x, y, z), where x = a2(a2 – b2)(c2 – a2), y = b2(a2 – b2)(b2 – c2), (4.5) z = c2(b2 – c2)(c2 – a2). It may now be checked that Q also lies on circles BVW and CWU. In similar fashion it may be shown that circles AWU, BUV, CVW all lie on the point R on the circumcircle with co-ordinates (x, y, z), where x = a2(a2 – b2(b2 – c2), y = b2 (b2 – c2)(c2 – a2), (4.6) 2 2 2 2 2 z = c (a – b )(c – a ). 5. Concluding remark Note that AVK, BWK, CUK are all straight lines, where K is the symmedian point, the first of these lines having equation b2x = a2y and so on. This is reminiscent of the fact that 4
AaHG, BbHG, CcHG are straight lines, see Article 19. It seems likely that other cases when ABC and UVW are in triple circular perspective are such that a point P exists on circle UVW such that AUP, BVP, CWP are straight lines.
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5
Article 34 Some Circles in the Cyclic Quadrilateral configuration Christopher J Bradley
E Q
A S
D
O
G
B T P R
C
F
Figure
1. Introduction When ABCD is a cyclic quadrilateral, centre O, and BA^CD = E, BC^AD = F and AC^BD = G, then circles ABF, CDF, BCE, ADE meet at a point T on EF. The perpendicular from T to EF passes through G and O. The centres of the four circles, O and T lie on a circle. These facts are established using Cartesian co-ordinates. These are not new results, though until now I did not know that the last mentioned circle passes through T as well as the five circle centres. See the figure above. 2. Preliminary analysis and the diagonal points E, F, G, and T We take ABCD to be the unit circle, origin the centre O, and A with co-ordinates (2a/(1 + a2), (1 – a2)/(1 + a2)). B, C, D have similar co-ordinates but with parameters b, c, d instead of a. The chord AB has equation (a + b)x + (1 – ab)y = (1 + ab), (2.1) Other chords have similar equations with appropriate change of parameters. The chords AB and CD meet at E with co-ordinates (x, y), where x = 2(ab – cd)/(abc + abd – acd – bcd + a + b – c – d), y = – (abc + abd – acd – bcd – a – b + c + d)/( abc + abd – acd – bcd + a + b – c – d). (2.2) F = BC^AD has similar co-ordinates to those in Equation (2.1), but with a and c interchanged. G = AC^BD has similar co-ordinates to those in Equation (2.1), but with b and c interchanged. The equation of EF is found to be 2(ac – bd)x – (abc – abd +acd – bcd – a + b – c +d)y – abc + abd – acd + bcd – a + b – c + d) = 0.
(2.3)
The perpendicular to EF through O has equation 2(ac – bd)y + (abc – abd +acd – bcd – a + b – c +d)x = 0.
(2.4)
It may now be checked that this line passes through G. Indeed it is well known that O is the orthocentre of triangle EFG. The lines with Equations (2.3) and (2.4) meet at a point T with co-ordinates (m/k, n/k), where m = 2(a2c(b(c – d) + cd + 1) – a(b2d(c – d) + b(c2d + c(d2 + 1) + d) – c(c – d)) + bd(b(cd + 1) – c + d)), n = – (a(b(c – d) + cd + 1) – b(cd + 1) + c – d)(a(b(c – d) + cd – 1) + b(1 – cd) – c + d)(2.5) k = a2(b2(c – d)2 + 2b(c – d)(cd – 1) + c2(d2 + 4) – 2cd + 1) – 2a(b2(c – d)(cd – 1) + b(c2 + 1)(d2 + 1) + (c – d)(cd – 1)) + b2(c2d2 – 2cd + 4d2 + 1) + 2b(c – d) (cd – 1) + (c – d)2. 3. The four circles through T Every circle may be expressed in the form
x2 + y2 + 2gx + 2fy + k = 0, and since this may be written in the form (x + g)2 + (y + f)2 = g2 + f2 – k,
(3.1)
(3.2)
it represents a circle, centre (– g, – f) and radius √(g2 + f2 – k). In order to obtain the equation of a circle through three given points one can put the coordinates of those points in turn into Equation (3.1), solve the three resulting equations for f, g, k and insert those values back into the equation. To specify a circle one may write down its equation or specify its values of f, g, k. The latter is useful if the centre is what is required. The equation of circle ABF turns out to be (abc + bcd – abd – acd – a + b + c – d)(x2 + y2) + 2(a + b)(d – c)x + 2(c – d)(ab – 1)y + abc – abd + acd – bcd + a – b + c – d = 0. (3.3) It may now be verified that this circle passes through the point T. Similarly the circles CDF, BCE, ADE all pass through T. 4.
The six point circle PQRSOT
The points P, Q, R, S are the centres of circles ABF, BCE, CDF, ADE respectively. Their co-ordinates are: For P: For Q:
((a + b)(c – d), (1 – ab)(c – d))/(abc – abd – acd + bcd – a + b + c – d); ((b + c)(a – d), (1 – bc)(a – d))/(abc + abd – acd – bcd + a + b – c – d);
(4.1) (4.2)
For R:
((b – a)(c + d), (b – a)(1 – cd))/(abc – abd – acd + bcd – a + b + c – d);
(4.3)
For S:
((a + d)(b – c), (1 – ad)(b – c))/abc + abd – acd – bcd + a + b – c – d);
(4.4)
The equation of the circle that contains all these points and also O and T is (abc + abd – acd – bcd + a + b – c – d)(abc – abd – acd + bcd – a + b +c – d)(x2 + y2) + (a2(b2(c – d)(cd – 1) + b(1 + c2)(1 + d2) +(c – d)(cd – 1)) – a(b2 + 1)(c2 + 1)(d2 + 1) + b2(c – d)(cd – 1) + b(c2 + 1)(d2 + 1) + (c – d)(cd – 1))x + (a2(b2(c + d)(c – d) + c2(d2 + 2) + 1) – b2(c2d2 + 2d2 + 1) + c2 – d2)y = 0. (4.5)
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ARTICLE 35 Where Seven Circles meet Christopher J Bradley
A
W
Y
G U X
P V
H Z C
B
Q
Figure
1
1. Introduction The orthocentroidal circle S of triangle ABC on GH as diameter possesses an unusual property. If you draw circles BHC, CHA, AHB to meet S again at points X, Y, Z and circles BGC, CGA, AGB to meet S again at U, V, W then the following property holds: circles AYX, BZX, CXY, AVW, BWU, CUV all pass through a point Q on the circumcircle of ABC. The points X, Y, Z are known to be the same points on S as those points where the medians AG, BG, CG respectively meet S. 2. The circle S and the points X, Y, Z, and circles AYZ, BZX, CXY The equation of the orthocentroidal circle S, see Bradley and Smith [1], is (b2 + c2 – a2)x2 + (c2 + a2 – b2)y2 + (a2 + b2 – c2)z2 – a2yz – b2zx – c2xy = 0.
(2.1)
The co-ordinates of X, Y, Z, where AG, BG, CG meet the orthocentroidal circle are X(a2, b2 + c2 – a2, b2 + c2 – a2), Y(c2 + a2 – b2, b2, c2 + a2 – b2), Z(a2 + b2 – c2, a2 + b2 – c2, c2). Now that these co-ordinates are known the equation of circle AYZ may be computed and is c2(c2 + a2 – b2)(c2 – a2)y2 – b2(a2 + b2 – c2)(a2 – b2)z2 + (a6 – a4(b2 + c2) – a2(b4 – 3b2c2 + c4) + (b2 – c2)(b4 – c4))yz – b2c2(c2 – a2)zx + b2c2(a2 – b2)xy = 0. (2.2) The equations of circles BZX and CXY may now be written down by cyclic change of a, b, c and x, y, z. 3. The point Q where circles AYZ, BZX, CXY meet The circle AYZ with Equation (2.2) meets the circumcircle at A and again at Q with co-ordinates (x, y, z), where x = a2/{(b2 + c2 – a2)(b2 – c2)}, y = b2/{(c2 + a2 – b2)(c2 – a2)}, (3.1) 2 2 2 2 2 2 z = c /{(a + b – c )(a – b )}. The symmetry of these expressions ensures that circles BZX, CXY also pass through this point. 4. The points U, V, W, and circles AVW, BWU, CUV The co-ordinates of U, V, W, where circles BGC, CGA, AGB meet the orthocentroidal circle are found to be, for U (x, y, z) where x = 3a2(a2 + b2 – c2)(b2 + c2 – a2), y = 3a6 + a4(b2 – c2) + a2(b2 – 3c2)(c2 – b2) + (b2 + c2)(b2 – c2)2, (4.1) 2
z = (c2 + a2 – b2)(3a4 + 2a2(b2 – c2) – (b2 – c2)(b2 + c2)) and those for V and W follow by cyclic change of a, b, c and x, y, z. The equation of circle CUV can now be obtained and is b2(a2 + b2 + c2)(b2 + c2 – a2)(c2 – b2)x2 + a2(a2 + b2 + c2)(c2 + a2 – b2)(a2 – c2)y2 – a2(2a6 – 2a4(b2 + c2) – a2(b4 – 5b2c2 + 2c4) + (b2 – c2)(3b4 + b2c2 – 2c4))yz + b2(a2 + b2 – c2)(3a4 + a2(c2 – 4b2) + 2(b4 – c4))zx – (a2 + b2 – c2)(a6 – a4(b2 + c2) – a2(b4 – 3b2c2 + c4) + (b4 – c4)(b2 – c2))xy = 0.
(4.2)
It may now be checked that Q lies on this circle and by symmetry lies also on the circles AVW, BWU. CABRI indicates that given any circle Σ in the plane of a triangle ABC (other than the circumcircle) the there is one and only one pair of diametrically opposite points D, E lying on it such that circles BDC, CDA, ADB, BEC, CEA, AEB meet Σ in points X, Y, Z, U, V, W such that circles AYZ, BZX. CXY, AVW, BWU, CUV have a common point Q lying on the circumcircle. We can offer no proof of this result, but it may be regarded as a safe conjecture.
Reference 1. C.J.Bradley and G.C.Smith, The locations of triangle centres, Forum. Geom., 6 (2006) 57-70.
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3
Article 36 Where 7 Circles meet
Part 2
Christopher J Bradley
Z
A
U V E
Q
P W
C X
B
T Y
Figure 1
1. Introduction This article should be read as the last of a sequence of articles, starting with Article 19 and continuing through 30 – 33 and 35, on circular perspective. It deals with case when a triangle 1
ABC is in circular perspective with a pair of triangles both of which have vertices on a circle. What happens is that if a circle Σ is drawn, other than the circumcircle S and not through any vertex, and arbitrary points D and E are chosen on it, then if circles BCD, CAD, ABD are drawn to meet Σ again at X, Y, Z respectively and if circles BCE, CAE, ABE are drawn to meet Σ at U, V, W respectively, then the following happens: circles AYZ, BZX, CXY concur at a point T on S and circles AVW, BWU, CUV meet at a point T' on S. However, if E is chosen in a particular way then T and T' coincide. In fact for each choice of D on Σ, there is only one point E on Σ with the required property and when E is chosen appropriately, then UX, VY, WZ are concurrent. What we do first is to consider a limiting case when the circle Σ degenerates into a line m. This situation is shown in Figure 1. The line m is a transversal of ABC and does not pass through any vertex. The points X, Y, Z are the points where m meets BC, CA, AB respectively. Q is a fixed point not at a vertex. It then follows that circles AYZ, BZX, CXY concur at the Miquel point T, which in the case of a transversal is well known to lie on the circumcircle S. The points U, V, W are now defined as the intersections of circles BCE, CAE, ABE with m. It now turns out that circles AVW, BWU, CUV always concur at a point T' on S. But there is one and only one position of E which results in the point T' coinciding with T. This turns out to be when the circles XUQ, YVQ, ZWQ are coaxal. Interestingly E does not move from its favourable position on m when the point Q is altered to another position. The reader, by now, will no doubt have appreciated that the general problem is now proved by inverting Figure 1 through the point Q. The fixed point Q is relabelled D, the coaxal circles simply become the lines XU, YV, ZW. It is hoped that no confusion is caused by not putting stars on the points in the inverted figure. The analysis for the configuration of Figure 1 is carried out using areal co-ordinates with ABC as triangle of reference and it is by no means straightforward. Some geometers might regard it as horrendous, but my retort would then be to ask them to prove it otherwise. 2. The Miquel point T for triangle ABC and the points X, Y, Z We take the transversal to have equation lx + my + nz = 0.
(2.1)
It follows that X, Y, Z have co-ordinates X(0, n, – m), Y(– n, 0, l), Z(m, – l, 0). The circle AYZ has equation a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0, (2.2)
2
where u = 0, v = c2m/(l – m), w= b2n/(l – n). The line with equation ux + vy + wz = 0, with these values of u, v, w is the equation of the common chord of circle AYZ and the circumcircle S and therefore meets S at A and the point T with co-ordinates x = a2mn(l – m)(n – l), y = b2nl(l – m)(m – n), (2.3) 2 z = c lm(m – n)(n – l). The symmetry of these co-ordinates under cyclic change of x, y, z and a, b, c and l, m, n shows that T also lies on circles BZX, CXY and is therefore the Miquel point. 3. The points U, V, W when E is chosen at random on XYZ In what follows it is best to parameterize points on the transversal XYZ, with Equation (2.1), with a parameter s such that a general point on it has co-ordinates (smn, (1 – s)nl, – lm). We now choose E to have parameter t so that E has co-ordinates (tmn, (1 – t)nl, – lm). The circle EBC has an equation of the form of Equation (2.2) with v = w = 0 and u = {l(a2l(t – 1) – t(b2m + c2n(t – 1)))}/{t(l(m + n(t – 1)) – mnt)}.
(3.1)
This circle meets XYZ at points with parameter s, where s = t (the point E) and at U with s = {a2l(l(m + n(t – 1)) – mnt)}/{n(a2l(l – m)(t – 1) – mt(b2(l – m) + c2(n – 1)))}.
(3.2)
The circle EAB has an equation of the form of Equation (2.2), with u = v = 0 and w = {n(a2l(t – 1) – t(b2m + c2n(t – 1)))}/{mnt – l(m + n(t – 1))}.
(3.3)
This circle meets XYZ at points with parameter s, where s = t (the point E) and at W with s = {l(a2m(l – n) + (n – m)(b2m + c2n(t – 1)))}/{c2n(l(m + n(t – 1)) – mnt)}.
(3.4)
The circle ECA has an equation of the form of Equation (2.2), with u = w = 0 and v = {m(a2l(t – 1) – t(b2m + c2n(t – 1)))}/{(1 – t)(l(m + n(t – 1)) – mnt)}.
(3.5)
This circle meets XYZ at points with parameter s, where s = t(the point E) and at V with s = {l(a2n(l – m)(t – 1) + (m – n)(b2m + c2n(t – 1)))}/{n(a2l(l – m)(t – 1) + b2mt(m – 1) + c2l(m – n)(t – 1))}. (3.6) The co-ordinates of U, V, W now follow from Equations (3.2), (3.4) and (3.6). 4. The circles AVW, BWU, CUV and the point T'
3
We omit calculation of the equations of the circles AVW, BWU, CUV which though arduous is straightforward. We record the co-ordinates of the point T' where they are concurrent on the circumcircle. Its co-ordinates are (x, y, z), where x = – a2/{a2l(l(m – n(t – 1)2) + mnt(t – 2)) +mnt2(b2(l – m) + c2(n – 1))}, (4.1) 2 2 2 2 2 2 2 2 y = – b /{a ln(l – m)(t – 1) + b m(mnt – l(m + n)(t – 1))) + c ln(m – n)(t – 1) }, (4.2) z = c2/{a2lm(l – n) + b2lm(n – m) + c2n(mnt2 – l(m(2t – 1) + n(t – 1)2))}. (4.3) For the points T and T' to coincide the ratios of their co-ordinates must be equal. This turns out to be a quadratic equation in t with one root equal to l(n – m)/n(l – m). However this would put the point E at infinity, which is not of interest as U, V, W would then coincide with X, Y, Z. This leaves a unique value of t, which is t = {l(a2(l(m + n) – 2mn) + (n – m)(b2m – c2n))}/{n(a2l(l – m) + b2m(m – l) + c2(l(2m – n) – mn))}. (4.4) 5. The positions of E, U, V, W when T and T' coincide The value of t in Equation (4.4) immediately gives us the co-ordinates of E when it assumes its special position. These are (x, y, z), where x = a2(l(m + n) – 2mn) + (n – m)(b2m – c2n), y = a2l(n – l) + b2(l(m – 2n) + mn) + c2n(l – n), (5.1) 2 2 2 z = a l(m – l) + b m(l – m) + c (mn – l(2m – n)). The circle EAB has an equation of the form (2.2) with u = v = 0 and w = – {a4l2 – 2a2l(b2m + c2n) + b4m2 – 2b2c2mn + c4n2}/{2(a2l(l – m) + b2m(m – l) + c2(l(2m – n) – mn))}.
(5.2)
This circle meets XYZ at E and again at the special position of W with co-ordinates (x, y, z), where x = m(a2l – b2m + c2n), y = – l(a2l – b2m – c2n), (5.3) 2 z = – 2lmc . We now take the point Q to have co-ordinates Q(p, q, r). The circle QWZ has an equation of the form (2.2), where u, v, w are given by the equations a2qr + b2rp + c2pq + (p + q + r)(pu + qv + rw) = 0, (5.4) 2 (l – m)(lv – mu) – c lm = 0, (5.5) 4 2 3 2 2 2 2 2 2 2 a (c l m + l (l – m)(lv – mu)) – 2a l(b m(c lm + (l – m)(lv – mu)) + c (c lmn + l (nv – m(v + w)) + lm2(u + w) – m2nu)) + b4m2(c2lm + (l – m)(lv – mu)) – 2b2c2m(c2lmn + l2(m(v + u) – nv) – lm2(u + w) + m2nu) + c6lmn2 + c4(l(2m – n) – mn)(l(2mw – nv) – mnu) = 0. (5.6)
4
The circle EBC has the form of Equation (2.2) with v = w = 0 and u = {a4l2 – 2a2l(b2m + c2n) + b4m2 – 2b2c2mn + c4n2}/{2(a2(l(m + n) – 2mn) + (n – m)(b2m – c2n))}.
(5.7)
This circle meets XYZ at E and again at the special position of U with co-ordinates (x, y, z), where x = 2a2mn, y = – n(a2l + b2m – c2n), (5.8) 2 2 2 z = – m(a l – b m + c n). The circle QUX has an equation of the form (2.2), where u, v, w are given by the equations a2qr + b2rp + c2pq + (p + q + r)(pu + qv + rw) = 0, (5.9) 2 (m – n)(mw – nv) – a mn = 0, (5.10) 6 2 4 2 2 2 2 2 2 2 a l mn – a (2b lm n + 2c lmn + (2mn – l(m + n))(l(mw + nv) – 2mnu)) + a (b m – c n)(b2m2n – c2mn2 – 2(l(m2w – n2v) – mn(m(u + w) – n(u + v)))) + (m – n)(b2m – c2n)2(mw – nv) = 0.(5.11) In the same way it can be shown that circle ECA meets XYZ at E and again at the special position of V with co-ordinates (x, y, z), where x = – n(a2l + b2m – c2n), y = 2b2nl, (5.12) 2 2 2 z = – l(– a l + b m + c n). Once the values of u, v, w have been obtained from Equations (5.4) – (5.6) and (5.9) – (5.11) the equation of the common chord of the circles QWZ and QUX may be obtained as (ux + vy + wz)QWZ = (ux + vy + wz)QUX (5.13) In the same way the equation of the common chord of circles QWZ and QVY may be obtained and is found to coincide with the common chord having Equation (5.13). This proves that the condition for E to have special position on XYZ in order for T and T' to coincide the circles QUX, QVY, QWZ must be coaxal. Inverting with respect to Q proves the more general theorem when the eight points D, E, X, Y, Z, U, V, W lie on a circle and the condition then for T and T' to coincide is that UX, VY, WZ are concurrent. See Figure 2 below
5
A
Z U V
E P D Y
X
B
C W T
Figure 2
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Article 37 More Circles in the Cyclic Quadrilateral Configuration Christopher J Bradley
F
A T P
Y S
W X
O
B G Q
E
Z C R
D
Figure
1
1. Introduction In the figure ABCD is a cyclic quadrilateral, centre O, with AB^CD = E, AD^BC = F, AC^BD = G so that EFG is the diagonal point triangle. The midpoints of AB, BC, CD, DA, EF are P, Q, R, S, T respectively. Circles ASP, BPQ, CQR, DRS are drawn and they all pass through O. The circles ASP, CQR meet at O and Y, circles BPQ, DRS meet at O and Z. It is shown that circle OYZ passes through G and has centre W, the midpoint of OG. It follows that angles OYG, OZG are right angles. The lines PR, QS meet at a point X. We show that X is the midpoint of YZ and that points T, X, Y, Z are collinear. In the sections that follow we use Cartesian co-ordinates with circle ABCD as unit circle. 2. The four circles through O The point A has co-ordinates (2a/(1 + a2),(1 – a2)/(1 + a2)) and B, C, D have similar co-ordinates but with parameters b, c, d rather than a. The point P, the midpoint of AB, has co-ordinates (x, y), where x = {(a + b)(1 + ab)}/{(1 + a2)(1 + b2)}, (2.1) 2 2 y = {(1 – ab)(1 + ab)}/{(1 + a )(1 + b )}. (2.2) Points Q, R, S have similar co-ordinates but with parameters chosen appropriately. The equation of the circles ASP is (1 + a2)(x2 + y2) – 2ax – (1 – a2)y = 0. (2.3) Circles BPQ, CQR, DRS have similar equations, but with parameters b, c, d rather than a. It is easy to see that these circles all pass through O, and indeed that their radii are ½ and they lie on AO, BO, CO, DO as diameters. 3. The points Y, Z and the circle OYZG The circles ASP and CQR meet at O and at a point Y with co-ordinates as in Equations (2.1) and (2.2) but with c instead of b. This means that Y is the midpoint of AC. Similarly the circles BPQ and DRS meet at O and at Z, the midpoint of BD. The circle OYZ may now be computed and has equation (abc – abd + acd – bcd + a – b + c – d)(x2 + y2) + 2(bd – ac)x + (abc – abd + acd – bcd – a + b – c + d)y = 0.
(3.1)
The lines AC and BD meet at the point G with co-ordinates (x, y), where x = 2(ac – bd)/(abc – abd + acd – bcd + a – b + c – d), (3.2) y = (abc – abd + acd – bcd – a + b – c + d)/(abc – abd + acd – bcd + a – b + c – d). (3.3)
2
It may now be checked that G lies on the circle OYZ and indeed that the centre W of the circle OYZG is the midpoint of OG. It follows that angles OYG and OZG are each 90o. 4. The midpoint X of YZ is also the intersection of the diagonals of parallelogram PQRS From the co-ordinates of Y and Z the co-ordinates of X, the midpoint of YZ may be obtained and they are (x, y), where x = Σ a/{2(1 + a2)}, (4.1) 2 y = – 1 + Σ 1/{2(1 + a )}. (4.2) In Equations (4.1) and (4.2) the sum is taken over a, b, c, d. The co-ordinates of P are given by Equations (2.1) and (2.2) and those of R are similar, but with c, d instead of a, b. The midpoint of PR may now be computed and it is found to have the same co-ordinates as X. Similarly X is also the midpoint of QS. 5. The points E, F, T and the collinearity of T, Y, Z The point E is the intersection of AB and CD and has co-ordinates (x, y), where x = {2(ab – cd)}/{abc + abd – acd – bcd + a + b – c – d}, y = –{abc + abd – acd – bcd – a – b + c + d}/{abc + abd – acd – bcd + a + b – c – d}. The point F is the intersection of BC and AD and has co-ordinates (x, y), where x = {2(bc – ad)}/{abc – abd – acd + bcd – a + b + c – d}, y = – {abc – abd – acd + bcd + a – b – c + d}/{abc – abd – acd + bcd – a + b + c – d}.
(5.1) (5.2)
(5.3) (5.4)
The co-ordinates of Y and Z are known, so the equation of YZ is (a2b2c2 – a2b2d2 + a2c2d2 – b2c2d2 + 2a2c2 – 2b2d2 + a2 + c2 – b2 – d2)x + (abcd(abc – abd + acd – bcd) – a2b2c + a2b2d + a2bc2 + a2bd2 + a2c2d – a2cd2 – ab2c2 – ab2d2 – ac2d2 + b2c2d – b2cd2 + bc2d2 + a2b – a2c + a2d – ab2 – ac2 – ad2 – b2c + b2d + bc2 + bd2 + c2d – cd2 – a + b – c + d)y + (1 + ac)(1 + bd)(abc – abd – bcd + acd +a – b + c – d) = 0. (5.5) It may now be shown that T, the midpoint of EF lies on this line.
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Article 38 Eight points on a line and seven circles through a point Christopher J Bradley
A U T Z
W X
V D E
B
C
Figure
1. Introduction Triangle ABC and a transversal are given and any point D (not on the sides of ABC) is chosen on the transversal. Circles BCD, CAD, ABD meet the transversal again at points X, Y, Z. Circles AYZ, BZX, CXY are now drawn and they are found to concur on the circumcircle ABC at a point T. The process is repeated with a variable point E intead of D, yielding three further points U, V, W on the transversal. Circles AVW, BWU, CUV are now drawn and, in general, they meet at a point S on the circumcircle, distinct from T. However, it is shown there are two positions of 1
E when S coincides with T. One is obviously when E coincides with D and the other is at a point related to E by a formula that we discover. Bearing in mind the results of Articles 19 and 30-32 we may say that circular perspectives are found between ABC and the singular circles XYZ and UVW that have a common vertex of perspective. The analysis of the situation described above and illustrated in the figure is developed using areal co-ordinates with ABC as triangle of reference. 2. The point D, the circle BCD and the point X We take the transversal to have equation lx + my + nz = 0. Points on this line may be parameterized by p, with x = mnp, y = nl(1 – p), z = – lm. We take D to have parameter t, so its co-ordinates are D(mnt, nl(1 – t), – lm). The equation of any circle using areal co-ordinates may be put in the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
In order to find the equation of any particular circle through three given points, one substitutes the co-ordinates of the points into Equation (2.1) to provide three equations for u, v, w. For the circle BCD we find v = w = 0 and u = (a2l2(t – 1) – b2lmt – c2nlt(t – 1))/(lmt + nlt(t – 1) – mnt2).
(2.2)
The circle BCD now meets the transversal at a point with parameter s given by s = t (that is at D) and the point X with parameter s = a2(l2m + nl2(t – 1) – lmnt)/(a2ml(l – m)(t – 1) – b2mn(l – m)t – c2mn(n – l)t). (2.3) The co-ordinates of X may therefore be taken as (x, y, z), where x = a2(lm + nl(t – 1) – mnt), y = – (a2l(l – n) + b2n(l – m)t + c2n(n – l)t), z = – (a2l(l – m)(t – 1) – b2m(l – m)t – c2m(n – l)t.
(2.4)
3. The circle CAD and the point Y Following the same approach as in section 2 we obtain for circle CAD the values u = 0, w = 0, v = (a2lm(t – 1) – b2m2t – c2mnt(t – 1))/{(1 – t)(lm + nl(t – 1) – mnt)}, (3.1) and the co-ordinates of Y as (x, y, z), where x = a2n(l – m)(t – 1) + b2m(m – n) + c2n(m – n)(t – 1), y = b2(mnt – lm – nl(t – 1)), z = a2l(l – m)(1 – t) + b2m(l – m)t + c2l(m – n)(1 – t). 2
(3.2)
4. The circle ABD and the point Z In the same way we obtain for circle ABD the values u = 0, v = 0, w = (a2nl(t – 1) – b2mnt – c2n2t(t – 1))/(mnt – lm – nl(t – 1)), and the co-ordinates of Z as (x, y, z), where x = a2m(l – n) + b2m(n – m) + c2n(n – m)(t – 1), y = a2l(n – l) + b2l(m – n) + c2n(l – n)t, z = c2(mnt – lm – nl(t – 1)).
(4.1)
(4.2)
5. The circle AYZ and its intersection T with the circumcircle Finding u, v, w for the circle AYZ is similar only technically more complicated. We find u = 0, v = c2a2nl(l – m)(t – 1)2 + b2c2m2nt2 – b2c2lm2 – b2c2lmn(t2 – 1) + c4nl(m – n)(t – 1)2 all divided by l(1 – t)(a2(l – m)(l – n) + b2(n – m)(l – m) + c2(n – m)(n – l)), (5.1) 2 2 4 2 2 2 2 2 2 2 2 2 2 2 w = a b lm(l – n) + b lm(n – m) + b c mn t – 2b c lmnt + b c lmn + b c ln (t – 1)2 all divided by l(1 – t)(a2(l – m)(l – n) + b2(l – m)(n – m) + c2(l – n)(m – n)). (5.2) Since the equation of the circumcircle is a2yz + b2zx + c2xy = 0,
(5.3)
it follows that the common chord of circles AYZ and ABC has equation vy + wz = 0, where v and w are given by Equations (5.1) and (5.2). Solving simultaneously we find the common points of these two circles are A and a point T with co-ordinates (x, y, z), where x = – a2(a2lm(l – n) + b2lm(n – m) + c2mn2t2 – c2lmn(2t – 1) + c2n2l(t – 1)2) .(a2nl(l – m)(t – 1)2 + b2m2nt2 – b2lm2 – b2lmn(t – 1)2 + c2nl(m – n)(t – 1)2)), y = – b2(a2lm(l – n) + b2lm(n – m) + c2mn2t2 – c2lmn(2t – 1) + c2n2l(t – 1)2) .(a2l2m – a2nl2(t – 1)2 + a2lmnt(t – 2) + b2mn(l – m)t2 + c2mn(n – l)t2), (5.4) 2 2 2 2 2 2 2 2 2 2 2 z = c (a l m – a nl (t – 1) + a lmnt(t – 2) + b mn(l – m)t + c mn(n – l)t ) .(a2nl(l – m)(t – 1)2 + b2m2nt2 – b2lm2 – b2lmn(t – 1)2 + c2nl(m – n)(t – 1)2)). It now follows from the general theory of circular perspective that circles BZX and CXY also pass through thus point. 6. What happens when another point E is chosen and the work repeated If we now choose a point E with parameter s, instead of D, the work follows as before, but with s replacing t, and Equations (5.4) produce a distinct point S on the circumcircle. However, if the ratios of the co-ordinates of S and T are equal it is found that there are four solutions for s, two of which are complex, s = t, of course, when E coincides with D, but there is another point E on the 3
transversal when circles AVW, BWU, CUV all pass through T. The value of s for which this occurs is s = (b2mt – c2nt – a2l(t – 2))/(a2l – b2m + c2n(1 – 2t)). (6.1) See the figure again, in which the point Y is off the paper.
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Article 39 The 60o, 75o, 45o Triangle and its Euler line Christopher J Bradley
P
B 75 R
H
O 60
45 C
A
Q
Figure
1. Introduction Let ABC be a 60o, 75o, 45o triangle and suppose that the Euler line OH, where O is the circumcentre and H the orthocentre, meets the sides BC, CA, AB at P, Q, R respectively. We prove that PR = RQ and RH = OQ. It is also the case that AQ = AR so that angle BPR = 15o. 2. The sides of the triangle and u, v, w 1
The sides are proportional to the sines of the angles of the triangle and may therefore be taken as a = ½√3, b = ¼(√6 + √2), c = ½√2. In the sections that follow we use the co-ordinates of H(u, v, w), where u = 1/(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2). (2.1) 3. The Euler line and the points P, Q, R The Euler line passes through H(u, v, w) and O(v + w, w + u, u + v) and so has the equation (v – w)x + (w – u)y + (u – v)z = 0. (3.1) The Euler line meets BC, x = 0, at P(0, 1 + (1/3)√3, – (1/3)√3). It meets CA, y = 0 at Q(1 – (1/3)√3, 0, (1/3)√3), and it meets AB, z = 0 at R(½ – (1/6)√3, (1/6)√3 + ½, 0). It is clear that PR = RQ, since xP + xQ = 2xR. 4. O, H and RH = OQ The values of u, v, w given in Equation (2.1), which when normalized, are the co-ordinates of H(2 – √3, (1/3)√3, (2/3)√3 – 1) and those of O are (½√3 – ½ , ½ –(1/6)√3, 1 – (1/3)√3). It now follows that xH + xO = xQ + xR so that RH = OQ. From the co-ordinates of Q and R and the side lengths b and c it immediately follows that triangle RQA is equilateral and hence that angle BPR = 15o. 5. The circumcircle of ABC, Circle CQH and the point U The equation of the circumcircle, using u, v, w, is u(v + w)yz + v(w + u)zx + w(u + v)xy = 0.
(5.1)
This becomes (9√3 – 15)yz + (√3 – 1)zx + (6√3 – 10)xy = 0. The equation of the circle CQH is (√3 – 1)x2 + (33 – 19√3)y2 + 2(3 – 2√3)yz + 2(√3 – 2)zx + 2(7– 4√3)xy = 0. These two circles meet at C and again at a point U with co-ordinates (1/13)(7 – 5√3), (11//3)√3 + 7, (4/3)√3 – 1). The point U is a key point in this configuration.
2
(5.2)
(5.3)
6. The circles BRH, COQ and the points S and T The equation of the circle BRH is 2(√3 – 2)x2 + 2(5√3 – 9)z2 + (9 – 5√3)yz + (11√3 – 19)zx + 2(7 – 4√3)xy = 0.
(6.1)
It may now be verified that this circle passes through U. Also it meets BC at B and the point S with co-ordinates S(0, 2/3, 1/3). That is, where SC = 2BS. This is a pleasing and unexpected result. The circle COQ has equation (√3 – 1)x2 + (9 – 5√3)y2 + 2(5√3 – 9)yz + 2(√3 – 2)zx + 2(3√3 – 5)xy = 0.
(6.2)
It may now be verified that this circle passes through the point S. It also meets the circumcircle at a point T with co-ordinates (1/6)(3 – 3√3, √3 + 3, 2√3). The points T, S, H turn out to be collinear.
P
U
B 75 R
T S H
O 60
45 C
A
Q
Figure
7. Circles AOU and PUQ pass through S The equation of circle AOU is 3
2(7 – 4√3)y2 – 2(√3 – 1)z2 + (17√3 – 29)yz + (√3 – 1)zx + 2(5√3 – 8)xy = 0.
(7.1)
It may now be verified that this circle passes through S. The equation of circle PUQ is (√3 – 1)x2 + (3√3 – 5)y2 + 4(√3 – 2)z2 + 2(7 – 4√3)yz + 2(2√3 – 3)zx + 6(2 – √3)xy = 0.
(7.2)
And this circle also passes through S. 8. Circle ABH passes through P The equation of circle ABH is 2(3 – 2√3)z2 + (5√3 – 9)yz + (5 – 3√3)zx + 2(3√3 – 5)xy = 0. It may now be verified that this circle passes through P. Finally it may now be verified that circles ABH and PUQ touch at P.
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(8.1)
Article 40 The 120o, 45o, 15o Triangle and its Euler Line Christopher J Bradley
C 30 30 15 15 15
O
75 30 15
P 105 60 S
60
U Q 90 T
60 R
120 30
A
120 135
45 30
30
15 15
Figure H
1
B
1. Introduction We find that the triangle with angles A, B, C equal to 120o, 45o, 15o respectively together with its Euler line OH, where O is the circumcentre and H the orthocentre, exhibits a number of interesting results. We define P, Q, R to be the points OH^BC, OH^CA, OH^AB respectively. We prove in what follows the following results: (i) PQ = QR; (ii) OQ = RH; (iii)Triangle AQR is equilateral; (iv) If U is the fourth vertex of parallelogram RAQU, then U lies on the circumcircle of ABC; (v) Circle PAC passes through H; (vi) Circle CBH passes through O; (vii)Circle COQ passes through U; (viii)Circle COQ meets BC again at a point S such that CS = 2SB; (ix)Circle OBR passes through U; (x) Circle APR passes through U and S. As a consequence of these results and angle-chasing angles between various lines are as shown in the figure above. In the following areal-co-ordinates are used with u, v, w instead of a, b, c, where u = 1//(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2) (1.1) and ABC as triangle of reference. 2. Values of a, b, c and the co-ordinates of H and O The sides of ABC are proportional to the Sines of 120o, 45o, 15o so we set a = ½√3, b = ½√2 and c = ¼(√6 – √2) Then, from Equations (1.1) we find, after normalization so that u + v + w = 1, that u = 2 + √3, v = – 1 – (2√3)/3, w = – (√3)/3. (2.1) These are now the normalized co-ordinates of the orthocentre, H. The co-ordinates of the circumcentre O are now ½(v + w, w + u, u + v) = (x, y, z) so that x = –½(1 + √3), y = 1 + (1/3)√3, z = (1/6)(3 + √3).
(2.2)
3. The Euler line and P, Q, R The equation of the Euler line through O, H is known to be (v – w)x + (w – u)y + (u – v)z = 0, which in this case is 2
(3.1)
(1 + √3)x + (4 + 2√3)y – (5 + 3√3)z = 0.
(3.2)
The co-ordinates of P, Q, R, where the Euler line meets BC, CA, AB respectively are P(0, (1/3)√3, 1 – (1/3)√3), Q((1/6)(3 + √3), 0, (1/6)(3 – √3)), R(1 + (1/3)√3, – (1/3)√3, 0). (3.3) 4. Results (i) – (iv) and the point U From the co-ordinates of P, Q, R it may be seen that P + R = 2Q, so that PQ = QR. Similarly O + H = R + Q, so that OQ = RH. From the co-ordinates of A, Q, R we find AQ = AR = (1/12)(3√2 – √6) so that triangle AQR is equilateral. Since U is defined as the fourth vertex of the parallelogram RAQU, its co-ordinates are given by U = Q + R – A. that is U((1/2)(1 + √3), – (1/3)√3, (1/6)(3 – √3)). (4.1) It may now be checked that U lies on the circumcircle of triangle ABC, whose equation is 3(1 + √3)yz + 2(1 + √3)zx + (√3 – 1)xy = 0. (4.2) 5. Results (v) – (viii) and the point S Circle PAC has equation 2√3y2 – (3 + √3)yz – 2(1 + √3)zx + (1 + √3)xy = 0.
(5.1)
It may now be checked that H lies on this circle. Circle CBH has equation 2x2 + 3(1 + √3)yz + 2(2 + √3)zx + (1 + √3)xy = 0.
(5.2)
It may now be checked that O lies on this circle. Circle COQ has equation 2x2 + (3 + √3)y2 – 2(3 + √3)yz – 2(2 + √3)zx + 2(1 + √3)xy = 0.
(5.3)
It may now be checked that U lies on this circle. Also the circle COQ meets BC, x = 0, at the point S with co-ordinates (0, 2/3, 1/3) so that CS = 2BS, a rather unexpected trisection of the side BC. 6. Results (ix) and (x) Circle BOR has equation 3
(√3 – 1)x2 – (3 + √3)z2 + 2(3 + √3)yz + 2(1 + √3)zx + 2xy = 0.
(6.1)
It may now be checked that U lies on this circle. Circle APR has equation 2y2 + 2(1 + √3)z2 – (5 + √3)yz – 4zx + (√3 – 1)xy = 0. It may now be checked that both U and S lie on this circle.
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(6.2)
Article 41 The 105o, 60o, 15o Triangle and its Euler Line and Circles Christopher J Bradley
B
30
15 15
O
U
60
120 R
45 P
105
60 Q
C
H
Figure
1. Introduction
1
60 A
In the above figure we display the 105o, 60o, 15o triangle and its Euler line OH, where O is the circumcentre and H is the orthocentre, together with several circles, the square UOAH and several other features. In the following sections we prove the following results for this configuration: (i) If the Euler line meets the sides BC, CA, AB respectively at points P, Q, R then HQ = RO; (ii) The co-ordinates of U are obtained, where AP = PU; (iii) U lies on the circumcircle ABC; (iv) Circle BCH passes through O; (v) UOAH is a square; (vi) Triangle BUO is equilateral. Areal co-ordinates are used throughout with ABC as triangle of reference. The figure also contains other details and some angles, which can easily be established from the data and the above results. 2. The sides a, b, c and the co-ordinates of H and O The side lengths a, b, c may be chosen equal to Sin A, Sin B, Sin C respectively, and so a = ½√3, b = ¼(√6 – √2), c = ¼(√6 + √2).
(2.1)
The co-ordinates of H are (u, v, w), where u = 1/(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2) and after normalization these are u = – 1, v = 1 – (2/3)√3, w = 1 + (2/3)√3. (2.2) The co-ordinates of O are ½(v + w, w + u, u + v) = (1, (1/3)√3, – (1/3)√3).
(2.3)
The equation of the Euler line OH may now be determined and is 2x – (√3 + 1)y + (√3 – 1)z = 0.
(2.4)
3. The co-ordinates of P, Q, R and results (i) and (ii) Solving Equation (2.4) with x = 0, we obtain P, whose co-ordinates are therefore x = 0, y = (1/6)(3 – √3), z = (1/6)(3 + √3).
(3.1)
Solving Equation (2.4) with y = 0, we obtain Q, whose co-ordinates are therefore x = – (1/3)√3, y = 0, z = 1 + (1/3)√3.
(3.2)
Solving Equation (2.4) with z = 0, we obtain R, whose co-ordinates are therefore 2
x = (1/3)√3, y = 1 – (1/3)√3, z = 0.
(3.3)
Since the co-ordinates of H, Q, R, O satisfy H + O = Q + R, HQ = RO, which is result (i). If U is on AP and is such that P is the midpoint of AU, then the co-ordinates of U satisfy U = 2P – A and are therefore x = – 1, y = 1 – (1/3)√3, z = 1 + (1/3)√3. (3.4) 4. Results (iii) and (iv) The equation of the circumcircle ABC is u(v + w)yz + v(w + u)zx + w(u + v)xy = 0,
(4.1)
which becomes 3yz + (2 – √3)zx + (2 + √3)xy = 0.
(4.2)
It may now be checked that U lies on the circumcircle. The equation of the circle BCH is x2 – 3yz + (√3 – 1)zx – (√3 + 1)xy = 0.
(4.3)
It may now be checked that O lies on this circle. This is always the case when angle A = 60o. 5. Results (v) and (vi) We already know that P is the midpoint of both OH and AU, so UOAH is a parallelogram. We next show that AO = AH. The displacement AO is (0, (1/3)√3, – (1/3)√3) and the displacement AH is (– 2, 1 – (2/3)√3, 1 + (2/3)√3). Now given a displacement d = (f, g, h), the square of its length is known to be d2 = – a2gh – b2hf – c2fg. From equation (2.1) we have a2 = ¾, b2 = ¼(2 – √3), c2 = ¼(2 + √3). (5.1) It soon follows that AO2 = AH2 = ¼, so UOAH is a rhombus. Finally the displacement HO = (2, √3 – 1, – √3 – 1), so that HO2 = ½. Thus angle OAH = 90o and UOAH is a square. It is now easy to show BO2 = BU2 = ¼ and hence triangle BOU is equilateral. Other features indicated by Cabri II plus are that U is the centre of a circle passing through R and Q; R is the centre of a circle passing through U, A and Q; and R is the centre of a circle passing through B and H It soon follows that RUQ is an equilateral triangle as well as RQA.
3
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4
Article 42 A Cyclic Quadrilateral, 29 Points and 33 Lines Christopher J Bradley
ac
To ac
ab
A
T
Sc Y
ad
G
F Sa Q C
Sb
D
ESd U W V
B
S
bc
P R
cd bd
X To Z To H
Z
1. Introduction In the Figure ABCD is a cyclic quadrilateral, Sa, Sb, Sc, Sd are the symmedian points of triangles BCD, ACD, ABD, ABC respectively, ab is the intersection of the tangents to circle ABCD at A 1
and B, with ac, ad, bc, bd, cd similarly defined. By the definition of a symmedian point Sa is the point of concurrency of C bd, B cd and D bc and similarly for Sb, Sc, Sd. A catalogue of the other 15 points in the figure is: P = AB^CD, Z = AD^BC, U = AC^BD (so UZP is the diagonal point triangle of ABCD), Q = ASc^BSd, R = ASd^CSb, S = BSc^DSa, T = CSa^DSb, V = CSd^BSa, W = CSa^BSd, X = ASb^CSd, Y = ASb^DSc, E = BSc^DSa, F = DSa^CSb, G = DSb^ASc, H = BSa^DSc. It is proved in the following sections that these 29 points lie severally on 33 lines as follows: ABP, CDP, SaSbP, ScSdP, ad bc P, ac bd P are 6 lines concurrent at P; ACU, BDU, SaScU, SbSdU, ad bc U, ab cd U are 6 lines concurrent at U; ADZ, BCZ, SbScZ, SaSdZ, ab cd Z, ac bd Z are 6 lines concurrent at Z; The diagonal line ZP contains the 6 additional points H, S, R, X, bd, ac; The diagonal line PU contains the 6 additional points Y, G, W, V, bc, ad; The diagonal line UZ contains the 6 additional points E, Q, F, T, cd, ab; There are 12 additional lines all of the same type containing 5 points each: ASbXYcd, BScESad, CSdVXab, DSaFSbc, ASdERbc, BSaVHcd, CSbFRad, DScYHab, AScGQbd, BSdWQac, CSaWTbd, DSbGTac. Each point considered lies on at least three lines. Proof s are established using Cartesian coordinates, making full use of symmetry. 2. The Symmedian Points We take ABCD to be the unit circle with equation x2 + y2 = 1 and A to have co-ordinates A(2a/(1 + a2), (1 – a2)/(1 + a2)), with B, C, D having similar co-ordinates, but with parameters b, c, d instead of a. It is well known that chord AB has equation (a + b)x + (1 – ab)y = (1 + ab), (2.1) and that the tangent at A has equation 2ax + (1 – a2)y = (1 + a2).
(2.2)
Other chords and tangents of ABCD may be written down with suitable change of parameters. The tangents at A and B meet at the point ab with co-ordinates ((a + b)/(1 + ab), (1 – ab)/(1 + ab)). The co-ordinates of ac, ad, bc, bd, cd may also be written down by suitable change of parameters. The line ad B has equation 2(ad – b2)x + (b2(a + d) – 2abd + a + d – 2b)y + b2(a + d) – 2abd – a – d + 2b = 0.
2
(2.3)
The line bd A has the form of Equation (2.3) but with a and b interchanged. As is known these lines meet at the symmedian point Sc of triangle ABD, which therefore has co-ordinates (x, y), where x = (1/k)(a2(b + d) + a(b2 – 6bd + d2) + bd(b + d)), y = (1/k)(a2(b2 – bd + d2 – 1) – a(b2d + bd2 – b – d) + b2d2 – b2 + bd – d2), (2.4) 2 2 2 2 2 2 2 2 2 k = a (b – bd + d + 1) – a(b d + bd + b + d) + b d + b – bd + d . The co-ordinates of other symmedian points may be obtained by appropriate change of parameters. 3. The diagonal point triangle The chords AB and CD meet at the point P with co-ordinates 0028, y), where x = (2/h)(ab – cd), y = – (1/h)(abc + abd – acd – bcd – a – b + c + d), h = (abc + abd – acd – bcd + a + b – c – d).
(2.5)
When the equation of the line ScSd has been obtained it may be checked that it passes through P. Similarly SaSb passes through P. It follows by symmetry that ABCD and SaSbScSd have the same diagonal point triangle PUZ, where U = AC^BD and Z = AD^BC. The co-ordinates of U and Z follow from Equation (2.5) by suitable change of parameters. The equation of the diagonal line ZP may now be obtained and is 2(ac – bd)x – (abc – abd + acd – bcd – a + b – c + d)y + (– abc + abd – acd + bcd – a + b – c + d) = 0. (2.6) The equations of the other diagonal lines may now be obtained by suitable change of parameters. 4. Other points on ZP It may now be checked that both bd and ac lie on ZP. The equation of A cd is 2(a2 – cd)x – ((a2 + 1)(c + d) – 2a(1 + cd))y – ((a2 + 1)(c + d) + 2a(cd – 1)) = 0.
(4.1)
It may be checked that Sb lies on this line. The equation of C ab may be found from Equation (4.1) by interchanging a and c and also b and d. The lines A cd and C ab meet at the point X with co-ordinates (x, y), where x = (1/k)(a2(c + d) + a(c2 – 3c(b + d) + bd) + bc(c + d)), y = – (1/k)(a2(b(c – d) – c2 + 1) + a(b + c)(cd – 1) + bd(1 – c2) + c(c – d)), (4.2) 3
k = a2(b(c – d) – c2 – 1) + a(b + c)(cd + 1) – bd(1 + c2) + c(c – d). It may now be checked that X lies on ZP. The disposition of the remaining points Q, R, S, T, V, W, Y, E, F, G, H on the diagonal lines now follows by suitable change of parameters.
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Article 43 The Altitudes and Radii of a triangle and its circumcircle Christopher J Bradley
A
ca
ab
R cb V
P
O N
ba B
Q
bc
H ac C
1. Introduction In the figure a scalene triangle ABC is shown, with its altitudes AH, BH, CH and the radii AO, BO, CO of its circumcircle. The point ab is the intersection of AH and BO, with similar definitions of points ac, bc, ba, ca, cb. The point P is the midpoint of the points ab, ac lying on
the altitude AH. Points Q and R are similarly defined on BH and CH respectively. In this article we show that circle PQR is the circle on OH as diameter and circles ab bc ca and ba ac cb both pass through H. The first circle has centre N, the nine-point centre, and the centres of the other two circles are collinear with N and equidistant from N. Cartesian co-ordinates are used to establish these results, with circle ABC as the unit circle. 2. Lines AH, BH, CH, AO, BO, CO and the point H The vertex A is taken to have co-ordinates (2a/(1 + a2), (1 – a2/(1 + a2)) and B and C have similar co-ordinates with parameters b, c rather than a. The chord BC has equation (b + c)x + (1 – bc)y = (1 + bc). (2.1) The line perpendicular to this through A is the line AH, which has equation (1 + a2)(1 – bc)x – (1 + a2)(b + c)y = a2(b + c) + 2a(1 – bc) – b – c.
(2.2)
BH and CH have similar equations with appropriate cyclic change of co-ordinates. Their intersection H has co-ordinates (x, y), where x = (2/k)(a2(b + c)(1 + bc) + a(1 + b2)(1 + c2) + (1 + bc)(b + c)), y = – (1/k)(a2(b2(3c2 + 1) + c2 – 1) +b2(c2 – 1) – c2 – 3), (2.3) where k = (1 + a2)(1 + b2)(1 + c2). (2.4) The equation of AO is (1 – a2)x = 2ay, with similar equations for BO and CO with parameters b and c respectively, rather than a. 3. The points ab and ac, the points P, Q, R The points ab and ac are the intersections with AH of the radii OA and OB respectively. Thus ab has co-ordinates (x, y), where x = 2bh, y = (1 – b2)h, where h = ((a2 – 1)(b + c) + 2a(1 – bc))/((1 + a2)(1 + b2)(b – c)). (3.1) The point ac has similar co-ordinates, but with b and c exchanged. The midpoint P of ab and ac has co-ordinates (f(1 – bc), – f(b + c)), where f = ((a2 – 1)(b + c) + 2a(1 – bc))/((1 + a2)(1 + b2(1 + c2)). (3.2) Point Q, R on BH, CH have similar co-ordinates with appropriately chosen cycic changes of a, b, c. 4. The circle PQR
The equation of the circle PQR may now be obtained and has equation (1 + a2)(1 + b2)(1 + c2)(x2 + y2) – 2((1 + a2) (1 + bc)(b + c) + a(1 + b2)(1 + c2))x + (a2(b2(3c2 + 1) + c2 – 1) +b2(c2 – 1) – c2 – 3)y = 0.
(4.1)
It is obvious that this circle passes through O and it may be checked that it also passes through H. Indeed from Equations (2.3) and (2.4) it may be observed that its centre is midway between O and H, the nine-point centre N. 5. The circle ab bc ca The co-ordinates of the three points ab, bc, ca follow from Equation (3.1) and so the equation of the circle through these points may be calculated. The working is technically difficult, and the equation is very lengthy, and we relied, as in most of the previous Articles, on the algebra computer package DERIVE, and the geometry computer package CABRI II plus for drawings and for forecasting results. The equation of the circle is (1 + a2(1 + b2) (1 + c2)(a – b)(b – c)(c – a)(x2 + y2) + 2(a4(b4c2 + b3c(1 – c2) + b2(1 + c2)2 + bc(1 – c2) – c2) – a3(1 + c2)(b4c + b3(1 + c2) + 2b2c – b(1 + c2) + c) + a2(b4(c4 + 2c2 – 1) + 2b3c(1 – c2) + 2b2(c2 + 1)(c2 – 1) + 2bc(1 – c2) + c4 – 2c2 – 1) + a(1 + c2)(b4c – b3(1 + c2) + 2b2c + b(1 + c2) + c) + b4c2 + b3c(1 – c2) – b2(1 + c2)2 + bc(1 – c2) – c2)x + 2(a4(b4c – 2b3c2 + b2c (1 + c2) + b(1 + c4) + c3) + a3(b4 – 2b2c2 + 1)(1 + c2) – a2(2b4c3 – b3(1 – c2)2 + 2b2c(1 + c2) – b(c4 – 2c2 + 1) + 2c) + a(1 + c2)(b4c2 – 2b2 + c2) + b4c + b3(1 + c4) + b2c(1 + c2) – 2bc2 + c3)y + (a(b2 – 2bc – 1) + b2c + 2b – c)(a(2bc – c2 + 1) + b(1 – c2) – 2c)(a2(b + c) + 2a(1 – bc) – b – c) = 0. (5.1) The circle ba ac cb has a similar equation with, say, a and c interchanged. It has been checked that both these circles pass through H, but the other assertions about their centres has not been checked, though Cabri is so accurate there is no doubt about them.
Flat 4 Terrill Court 12-14 Apsley Road Clifton BRISTOL BS8 2SP
Article 44 A natural Enlargement and a resulting Collineation Christopher J Bradley
A'
feu
A
Feu k g
B
G K
X H h
C
B' C'
1. Introduction Given a triangle ABC, draw the angle bisectors to meet its circumcircle and at those three points draw the tangents to form a triangle A'B'C', which, since the three points are the midpoints of the minor arcs BC, CA, AB, is a triangle similar to ABC. In fact it turns out that triangle A'B'C' is
an enlargement from a point X, the enlargement factor being R/r, the circumcircle of ABC being the incircle of A'B'C'. In this short article we find the point X, which has the property that if J is any triangle centre of ABC and j the corresponding triangle centre in A'B'C' then jJX is a straight line. In the Figure above the lines hHX, gGX, kKX are shown. Remarkably it turns out that the Feuerbach point Feu lies on hHX, so that in any triangle Feu, X and H are collinear. The Figure illustrates this and displays the nine-point circles and incircles which are tangent at feu and Feu. Of course, one can produce an enlargement of a triangle about any point you want, but as this particular enlargement arises naturally the point X is of special significance, and we call it the centre of centres. It is actually X56, the Exsimilicenter (or External Similitude Centre) [1] and so this collineation is already known. For this reason this article is kept very brief. 2. The equations of the sides of A'B'C' The equation of the line joining the circumcentre O and the midpoint L of BC is (b2 – c2)x + a2(y – z) = 0. This meets the circumcircle at the point with co-ordinates (– a2, b(b + c), c(b + c)). The tangent to the circumcircle at this point is the line B'C' with equation (b + c)2x + a2(y + z) = 0. (2.1) The equations of C'A' and A'B' follow by cyclic change of a, b, c. These lines intersect at The point A' with co-ordinates (a(a2 + ab + 2bc + ca), – b2(a + b – c), – c2(c + a – b)). 3. The co-ordinates of X56, the Exsimilicenter The equation of AA' is therefore c2(c + a – b)y = b2(a + b – c)z.
(3.1)
The three lines AA', BB', CC' meet at X56, the Exsimilicenter (or External Similitude Centre), whose co-ordinates are (a2/(b + c – a), b2/(c + a – b), c2/(a + b – c)). 4. The Collineation X4X11X56 The co-ordinates of Feuerbach’s point Feu or X11 are ((b + c – a)(b – c)2, (c + a – b)(c – a)2, (a + b – c)(a – b)2) and it is now easily shown that the orthocentre X4, Feuerbach’s point X11 and X56 are collinear.
Reference 1. C. Kimberling, “Major Centers of Triangles," Amer. Math. Monthly 104 (1997) 431-438.
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Article 45 On the Circles defining the Brocard Points Christopher J Bradley
A
Z' Y' R X O
P
V
' X' Q
Z B
H Y C
Figure 1
1. Introduction
1
Since writing Article 43 a number of further results have emerged in connection with its configuration. However, this article may be read independently, without recourse to any previous results. I am extremely grateful to David Monk [1] for pointing out the connection between the configuration in Article 43 and the circles defining the Brocard points. In Figure 1 we display a triangle ABC with circumcentre O and orthocentre H. The points X = CH^BO, Y = AH^CO, Z = BH^AO. The points X' = BH^CO, Y' = CH^AO, Z' = AH^BO. The point P is the midpoint of YZ', Q is the midpoint of ZX' and R is the midpoint of XY'. We prove the following results: (i) Circle PQR has OH as a diameter; (ii) Circle XYZ passes through H and Ω, the first Brocard point; (iii) Circle X'Y'Z' passes through H and Ω', the second Brocard point; (iv) Lines XX', YY', ZZ' are concurrent at a point V. (v) Triangle XYZ is similar to triangle CAB and triangle X'Y'Z' is similar to triangle BCA (where ordering of letters is important). In Figure 2 we display, in addition, circles BCX, CAY, ABZ, BCX', CAY', ABZ' and we prove the following results: (vi) Circles BCX, CAY, ABZ all pass through Ω; (vii) Circles BCX', CAY', ABZ' all pass through Ω'. Thus these circles are the defining circles of the Brocard points. So, circle ABZ touches BC at B, circle BCX touches CA at C, and circle CAY touches AB at A. And circle ABZ' touches CA at A, circle BCX' touches AB at B and circle CAY' touches BC at C. In Figure 3 we display, instead of the six circles in Figure 2, the following six circles: AYZ, BZX, CXY, AY'Z', BZ'X', CX'Y' and we prove the following result: (viii) All these six circles pass through the same point D lying on the circumference of the circumcircle of ABC. The point D is Tarry’s point, associated with the Brocard configuration, and has the following additional properties: (ix) AH^BD = E lies on circle CXY, BH^CD = F lies on circle AYZ, CH^AD = G lies on circle BZX, BH^AD = T lies on circle CX'Y', CH^BD = U lies on circle AY'Z' and AH^CD = S lies on circle BZ'X'. Apart from Results (i) and (v), the above results are proved using areal co-ordinates with ABC as triangle of reference. 2. The points P, Q, R, X, Y, Z, X', Y', Z' and the circle PQR 2
Result (i) BO and OC are equally inclined to AH, so triangle OYZ' is isosceles. OP is therefore at right angles to AH. Similarly OQ is at right angles to BH and OR is at right angles to CH. It follows that P, Q, R lie on the circle with OH as diameter. The co-ordinates of X, Y, Z, X', Y', Z' The equation of AH is (c2 + a2 – b2)y = (a2 + b2 – c2)z
(2.1)
c2(a2 + b2 – c2)x = a2(b2 + c2 – a2)z.
(2.2)
and the equation of BO is
These lines meet at the point Z' with co-ordinates (x, y, z), where x = a2(b2 + c2 – a2)(c2 + a2 – b2), y = c2(a2 + b2 – c2)2, z = c2(a2 + b2 – c2)(c2 + a2 – b2).
(2.3)
The co-ordinates of points X' and Y' follow from those of Z' by cyclic change of both a, b, c and x, y, z. The equation of CO is b2(c2 + a2 – b2)x = a2(b2 + c2 – a2)y. This line meets AH, with Equation (2.1) at the point Y with co-ordinates (x, y, z), where x = a2(b2 + c2 – a2)(a2 + b2 – c2), y = b2(c2 + a2 – b2)(a2 + b2 – c2), z = b2(c2 + a2 – b2)2.
(2.4)
(2.5)
The co-ordinates of points Z and X follow from those of Y by cyclic change of both a, b, c and x, y, z. 3. The circle XYZHΩ and result (ii) From the co-ordinates of X, Y, Z|, see Equation (2.5) we may obtain the equation of the circle XYZ, which is b2(b2 + c2 – a2)(c2 + a2 – b2)x2 + c2(c2 + a2 – b2)(a2 + b2 – c2)y2 + a2(a2 + b2 – c2)(b2 + c2 – a2)z2 + (b2 + c2 – a2)(a2b2 + b2c2 – 2c2a2 – b4)yz + (c2 + a2 – b2)(b2c2 + c2a2 – 2a2b2 – c4)zx + (a2 + b2 – c2)(c2a2 + a2b2 – 2b2c2 – a4)xy = 0. (3.1) 3
It may now be checked that both H and Ω(1/b2, 1/c2, 1/a2) lie on this circle. 4. The circle X'Y'Z'HΩ' and result (iii) From the co-ordinates of X', Y', Z', see Equation (2.3) we may obtain the equation of the circle X'Y'Z', which is c2(a2 + b2 – c2)(b2 + c2 – a2)x2 + a2(b2 + c2 – a2)(c2 + a2 – b2)y2 + b2(c2 + a2 – b2)(a2 +b2 – c2)z2 + (a2 + b2 – c2)(a2b2 + b2c2 – 2c2a2 – b4)yz + (b2 + c2 – a2)(b2c2 + c2a2 – 2a2b2 – c4)zx + (c2 + a2 – b2)(c2a2 + a2b2 – 2b2c2 – a4)xy = 0. (4.1) It may now be checked that both H and Ω'(1/c2, 1/a2, 1/b2) lie on this circle. 5. Result (iv) If one considers the point V with co-ordinates (x, y, z), where x = a2(b2 + c2 – a2)(a4b2 + a4c2 – 2a2b4 + 4a2b2c2 – 2a2c4 + b6 – b4c2 – b2c4 + c6), (5.1) and with y, z obtained by cyclic change of a, b, c, then it may now be verified that X, X', V are concyclic, that Y, Y', V are concyclic, and that Z, Z', V are concyclic. 6. Result (v) It is a simple piece of angle chasing that result (v) holds, namely triangle XYZ is similar to triangle CAB and triangle X'Y'Z' is similar to triangle BCA (where ordering of letters is important). In fact, it is true for any circle passing through H that the intersections of the circle with the altitudes through H form a triangle similar to ABC. 7. The six circles like AZB The equation of the circle AZB is easily calculated and is a2z2 + (a2 – b2)zx – c2xy = 0.
(7.1)
The exciting fact about this circle are that it is tangent to BC at B, and is therefore one of the defining circles of a Brocard point. And indeed it is easy to check that it passes through Ω, whose areal co-ordinates are (1/b2, 1/c2, 1/a2). In similar fashion it may be shown that circles BXC, CYA both pass through Ω, and that circles AZ'B, BX'C, CY'A all pass through Ω'(1/c2, 1/a2, 1/b2).
4
A
Z' Y' R X
O
P
V
' B
Q
Z
H
X' Y
C
Figure 2
8. The six circles like AYZ The equation of the circle AYZ may now be obtained and is c2(a2 + b2 – c2)(a4 – a2b2 – c2b2 + c4)y2 – c2(a2 + b2 – c2)(a4 – a2c2 – b2c2+ b4)z2 + (c2 – b2)(a6 – 2a4b2 – a4c2 + a2b4 + a2c4 + c2b4 – c6)yz – (a6c2 + a4(b4 – 2c4) + a2(c2 – b2)(2b4 + b2c2 + c4) + b4(b2 – c2)2)zx + c2(a6 – a4b2 + a2(b4 – c4) – b2(b2 – c2)2)xy = 0. (8.1)
5
G E
D A
U Z' R X V
Y' O
' Z
X'
Q
B
S T P H Y
F
C
Figure 3
This circle meets the circumcircle at A and the point D with co-ordinates (x, y, z), where x = 1/(b4 + c4 – a2(b2 + c2)), y = 1/(c4 + a4 – b2(c2 + a2)),
(8.2)
z = 1/(a4 + b4 – c2(a2 + b2). This point lies similarly on the five circles BZX, CXY, AY'Z', BZ'X', and CX'Y'. The point D is Tarry’s point, which is diametrically opposite the Steiner point, but in this configuration it has a further interesting property. See result (ix) above, which we now establish. The line AD has equation (c4 + a4 – b2(c2 + a2))y = (a4 + b4 – c2(a2 + b2))z. 6
(8.3)
The line AD meets AH with Equation (2.1) at the point G with co-ordinates (x, y, z), where x = (b2 – c2 – a2)(a4 + b4 – c2(a2 + b2)), y = (a2 – b2 – c2)(a4 + b4 – c2(a2 + b2)), (8.4) 2 2 2 4 4 2 2 2 z = (a – b – c )(c + a – b (c + a )). It may now be checked that G lies on circle BZX, which has an equation similar to Equation (8.1) with cyclic change of a, b, c and x, y, z. Similarly it may be proved that AH^BD = E lies on circle CXY, BH^CD = F lies on circle AYZ, BH^AD = T lies on circle CX'Y', CH^BD = U lies on circle AY'Z' and AH^CD = S lies on circle BZ'X'. Reference 1. D. Monk, private communication.
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7
A Mean of two Cevian Points and the Construction of Triangle Centres Christopher J Bradley
A
F
P D' R E
D
E'
Q F'
C
B
Figure
1.
Introduction
The idea of a triangle centre is well known. Over 3000 such points have been investigated and tabulated [1]. There are, of course, an infinity of triangle centres because every point on a line joining any two of them is also a triangle centre. For example, the Euler line collapses to the centroid when a triangle becomes equilateral. So the nine-point centre is a triangle centre, and one knows this without the need to work out its co-ordinates. Obviously there has to be some additional reason as to what actually qualifies as a triangle centre to make it worth tabulating. Presumably it must involve a construction of some kind relating it to the triangle or to other triangle centres that makes it significant or acceptable as an additional triangle centre. The aim of
this short article is to show that this is a very difficult task. My understanding is that Clark Kimberling decides when a point is significant. And the reason is not presumably that it lies on a line joining any two of them! What we do is to choose two points, produce a significant construction to provide a third point, which is a triangle centre if the two originally chosen ones are. Does the third point qualify or not? If the two chosen points are the centroid and the orthocentre the third point turns out to be the symmedian point. Does this convince one that the third point should qualify? If it does, remember there are already about 3000 triangle centres, so this would only add about 4.5 million. 2.
The construction
We choose any two points P and Q, not on the sides of the triangle and not so as to be on the same line through a vertex. Let P have co-ordinates (p, q, r) and Q have co-ordinates (u, v, w). AP has equation qz = ry and BQ has equation wx = uz. These lines meet at the point X(ur, wq, wr). BP^CQ = Y(up, vp, ru) and CP^AQ = Z(vp, vq, wq). Next we interchange P and Q and form X' = AQ^BP with co-ordinates (wp, vr, wr), BQ^CP = Y'(up, uq, wp), CQ^AP = Z'(uq, vq, vr). The equation of XX' may now be calculated and is found to be wr(wq – vr)x + wr(wp – ur)y + (r2uv – w2pq)z = 0.
(2.1)
The equation of YY' may be obtained from this by cyclic change of x, y, z and p, q, r and u, v, w. We now obtain the point of intersection R with co-ordinates x = up(wq + vr), y = vq(wp + ur), z = wr(vp + uq).
(2.2)
See the figure above. It is easy to see that if v, w › u and q, r › p , so that P, Q are triangle centres, then R is also a triangle centre. In particular if P is the centroid then R has co-ordinates (u(v + w), v(w + u), w(u + v)), and further if Q is the orthocentre, then R is the symmedian point. Reference 1. Clark Kimberling, "Major Centers of Triangles," Amer. Math. Monthly 104 (1997) 431-438.
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP
Article 47 A Cevian point and its Six Harmonics Christopher Bradley
C3 A
B3
N
C2
E A3
R F
W P
M
S
D
G A2
V U
K B
C1
B2
T J B1
L
C
L'
A1
Figure
1. Introduction Let P be a Cevian point of a triangle ABC and L, M, N the feet of the Cevians on BC, CA, AB respectively. Define the point B1 to be the harmonic conjugate of B with respect to L and C and let C1 be the harmonic conjugate of C with respect to B and L. Define points C2 and A2 on CA 1
and A3 and B3 on AB by a similar process involving harmonic conjugates. See the figure above. Now draw lines AB1, AC1, BC2, BA2, CA3 and CB3. The following additional Cevian points are thereby created: (i) R with Cevians ARL, BRC2, CRB3; (ii) S with Cevians ASB1, BSM, CSB3; (iii) T with Cevians ATB1, BTA2, CTN; (iv) U with Cevians AUL, BUA2, CUA3; (v) V with Cevians AVC1, BVM, CVA3; (vi) W with Cevians AWC1, BWC2, CWN. These six Cevian points we refer to as the harmonics of the initial Cevian point P. Six additional points D, E, F, G, J, K are also defined as follows: (i) D lying on BC2, A2A3, NL; (ii) E lying on AC1, B3B1, MN; (iii) F lying on AB1, C1C2, MN; (iv) G lying on CB3, A2A3, LM; (v) J lying on CA3, , B3B1, LM; (vi) K lying on BA2, C1C2, NL. Lines A3C1 and A2B1 are drawn and meet at A1. Similarly points B2, C3 are defined. The following results now hold: (i) A conic passes through A, B2, C, A1, B, C3; (ii) A conic passes through R, S, T, L, V, W; (iii) A conic passes through R, S, T, U, V, N; (iv) A conic passes through R, M, T, U, V, W. Also it is found that (i) R and U lie on APL; (ii) S and V lie on BPM; (iii) T and W lie on CPN. (iv) A conic passes through D, E, F, G, J, K In proving these assertions areal co-ordinates are used with ABC as triangle of reference and though full proofs would be prohibitively long and rather tiresome, what we have done is to give a catalogue of (i) sets of co-ordinates of each point; (ii) equations of all lines and conics. Checking results is then left to the reader. 2. Points B1, C1, C2, A2, A3, B3
2
Suppose P has co-ordinates (l, m, n). Then L has co-ordinates (0, m, n), M has co-ordinates (l, 0, n) and N has co-ordinates (l, m, 0). The conditions that {BB1; LC} = {CC1, LB} = – 1 mean that B1 has co-ordinates (0, m, 2n) and C1 has co-ordinates (0, 2m, n). Similarly C2 has co-ordinates (2l, 0, n) and A2 has co-ordinates (l, 0, 2n). And again A3 has co-ordinates (l, 2m, 0) and B3 has co-ordinates (2l, m, 0). 3. The Cevians AL, BM, CN: These have equations ny = mz, nx = lz, mx = ly respectively; AB1, AC1: These have equations 2ny = mz, ny = 2mz respectively; BC2, BA2: nx = 2lz, 2nx = lz respectively; CA3, CB3: 2mx = ly, mx = 2ly respectively. 4. The points A1, B2, C3 and the conic A1BC3AB2C A1 has co-ordinates (– l, 2m, 2z), B2 has co-ordinates (2l, – m, 2n) and C3 has co-ordinates (2l, 2m, – n). The conic A1BC3AB2C has equation lyz + mzx + nxy = 0. (4.1) Note that when P is the centroid this conic is the outer Steiner ellipse and when P is the symmedian point it is the circumcircle of ABC. 5. The points R, V, T, U, S, W R has co-ordinates (2l, m, n), V has co-ordinates (l, 2m, n) and T has co-ordinates (l, m, 2n); U has co-ordinates (l, 2m, 2n), S has co-ordinates (2l, m, 2n) and W has co-ordinates (2l, 2m, n). 6. The conics RSTLVW, VWRMTU, TUVNRS The conic RSTLVW has equation x2/l2 + 2y2/m2 + 2z2/n2 – 4yz/mn – zx/nl – xy/lm = 0.
(6.1)
The conic VWRMTU has equation 2x2/l2 + y2/m2 + 2z2/n2 – yz/mn – 4zx/nl – xy/lm = 0.
(6.2)
The conic TUVNRS has equation 2x2/l2 + 2y2/m2 + z2/n2 – yz/mn – zx/nl –4xy/lm = 0.
(6.3)
7. The points D, E, F, G, J, K
3
E'
It is found that D has co-ordinates (2l, 3m, n), that J has co-ordinates (l, 2m, 3n) and that F has co-ordinates (3l, m, 2n). Also it is found that E has co-ordinates 3l, 2m, n), that K has coordinates (l, 3m, 2n) and that G has co-ordinates (2l, m, 3n). Finally it is found that the conic DEFGJK has equation 11x2/l2 + 11y2/m2 + 11z2/n2 – 14yz/mn – 14zx/nl – 14xy/lm = 0.
(7.1)
It may be shown that B3C2, A3A2, MN are concurrent at L'(0, m, – n), the harmonic conjugate of L with respect to B and C. Similarly M' and N' on CA and AB respectively are harmonic conjugates of M and N respectively with respect to C and A and to A and B. If the points L, M, N are not the feet of Cevians, but are just the vertices of some arbitrary triangle, then many of the above results still hold. See the diagram below.
A C3
B3 E'' F C2
P A3
X W Y
F''
B2 Q VZ
B
E
U
C1
D
B1
R
D''
A1
Figure
4
A2
C
D'
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
5
Symmedian Point and its Harmonics Christopher J Bradley
To N'
C3 A
B3 N F L2
W Q
Y
E
K U R
L3
D C1
L
B2
V
N2 B
N1
P
A3
C2
X
M1
M3
M A2
Z C
B1
L'
A1 To M'
Figure
1. Introduction In the Figure above K is the Symmedian point and L, M, N are the feet of the Cevians through K. Lines MN, NL, LM are drawn. Points B1, C1 on BC are the harmonic conjugates of B in LC and C in BL and points C2, A2 and A3, B3 are defined similarly on CA and AB respectively. Lines AB1, AC1, BC2, BA2, CA3, CB3 are drawn. Points P = AL^MN, Q = BM^NL, R = CN^LM are shown, as well as U= BA2^CA3, V = CB3^AB1, W = AC1^ BC2.
1
It turns out that conics can be drawn passing through PQRWVL, PQRUWM, PQRVUN. Points X= B3A2^A3C2, Y = C1B3^B1A3, Z = A2C1^C2B1 are shown as well as D = A3B1^C1A2, E = B1C2^A2B3, F = C2A3^B3C1. It is then shown that conics may be drawn through DEFXYZ and C1B1A2C2B3A3. Points A1, B2, C3 are now located as A3C1^B1A2, B1A2^C2B3, C2B3^A3C1 respectively. It is then shown that A1, B2, C3 lie on the circumcircle of triangle ABC. Finally points M1 = AC1^MN, N2 = BA2^NL, L3 = CB3^LM, N1 = AB1^MN, L2 = BC2^NL, M3 = CA3^LM and a conic is shown to pass through L2L3M3M1N1N2. Working is carried out using areal co-ordinates with ABC as triangle of reference. Results are given, but not working as points, lines, and conics are so numerous. 2. Points K, L, M, N, B1, C1, A2, C2, B3, A3 The co-ordinates of these points are K(a2, b2, c2), L(0, b2, c2), M(a2, 0, c2), N(a2, b2, 0). The harmonic conjugate points have co-ordinates B1(0, b2, 2c2), C1(0, 2b2, c2), A2(a2 0, 2c2), C2(2a2, 0, c2), B3(2a2, b2, 0), A3(a2, 2b2, 0). Equations of various lines involving these points are: LM x/a2, + y/b2 – z/c2 = 0, MN y/b2 + z/c2 – y/b2 = 0, NL z/c2 + x/a2 – y/b2. AL c2y = b2z, BM a2z = c2x, CN b2x = a2y. AB1 2c2y = b2z, BC2 2a2z = c2x, CA3 2b2 x = a2y, AC1 c2y = 2b2z, BA2 a2z = 2c2x, CB3 b2x = 2a2y. The equation of the conic C1B1A2C2B3A3 is 2x2/a4 + 2y2/b4 + 2z2/c4 – 5yz/b2c2 – 5zx/c2a2 – 5xy/a2b2 = 0.
(2.1)
3. Points P, Q, R, U, V, W and conics PQRLVW, PQRMWU, PQRNUV The co-ordinates of the points are P = AL^MN (2a2, b2, c2), Q = BM^NL (a2, 2b2, c2), R = (a2, b2, 2c2), U = BA2^CA3 (a2, 2b2, 2c2), V = CB3^AB1 (2a2, b2, 2c2), W = AC1^BC2 (2a2, 2b2, c2). The equation of the conic PQRLVW is x2/a4 + 2y2/b4 + 2z2/c4 – 4yz/b2c2 – zx/c2a2 – xy/a2b2 = 0.
(3.1)
The equation of the conic PQRMWU is 2x2/a4 + y2/b4 + 2z2/c4 – yz/b2c2 – 4zx/c2a2 – xy/a2b2 = 0.
(3.2)
The equation of the conic PQRNUV is 2x2/a4 + 2y2/b4 + z2/c4 – yz/b2c2 – zx/c2a2 – 4xy/a2b2 = 0.
(3.3)
2
4. Points X, Y, Z, D, E, F and the conic through these points The co-ordinates of the points are X = B3A2^A3C2 (5a2, 2b2, 2c2), Y = C1B3^B1A3 (2a2, 5b2, 2c2), Z = A2C1^C2B1 (2a2, 2b2, 5c2), D = A3B1^C1A2 (a2, 4b2, 4c2), E = B1C2^ A2B3 (4a2, b2, 4c2), F = C2A3^B3C1 (4a2, 4b2, c2). The equation of the conic XYZDEF is 8x2/a4 + 8y2/b4 + 8z2/c4 – 11yz/b2c2 – 11zx/c2a2 – 11xy/a2b2 = 0.
(4.1)
5. The points A1, B2, C3 A1 = A3C1^B1A2,B2 = B1A2^C2B3, C3 = C2B3^A3C1. These points have co-ordinates (– a2, 2b2, 2c2), (2a2, – b2, 2c2), (2a2, 2b2, – c2) respectively. It may be checked that these points lie on the circumcircle of triangle ABC. 6. The points M1, N2, L3, N1, M2, L3 and the conic through these points These points have co-ordinates M1 (3a2, 2b2, c2), N2 (a2, 3b2, 2c2), L3 (2a2, b2, 3c2), N1 (3a2, b2, 2c2), L2 (2a2, 3b2, c2), and M3 (a2, 2b2, 3c2). It may be shown that N1 is the harmonic conjugate of N with respect to P and M and M1 the harmonic conjugate of M with respect to N and P, and similarly for other point pairs on NL and LM. The equation of the conic through M1, N2, L3, N1, M2, L3 is 11x2/a4 + 11y2/b4 + 11z2/c4 – 14yz/b2c2 – 14zx/c2a2 – 14xy/a2b2 = 0.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
3
(6.1)
4
Perpendicular Bisectors and Angle Bisectors Christopher Bradley
A
X
Z'' W
O
Y'
Y M
N
J
Q
V
U
X''
G I
Z
K
Z' X'
B
Y''
'
C
L
Figure 1
1. Introduction
1
In triangle ABC the incentre is I and the circumcentre is O. The angle bisectors AI, BI, CI are drawn, as well as the perpendicular bisectors of the sides OL, OM, ON. Of the nine intersections of these pairs of three lines, three, namely AI^OL, BI^OM, CI^ON are on the circumcircle (at the midpoints of the minor arcs). The other six are labeled as follows AI^OM = Y, BI^ON = Z, CI^OL = X, OL^BI = X', OM^CI = Y', ON^AI = Z'. Circles XYZ and X'Y'Z' are drawn, and for reasons that will shortly become clear the 7-point circle is also drawn. The following results now hold: (i) Circles XYZ and X'Y'Z' both pass through I, and they pass through the Brocard points Ω and Ω' respectively; (ii) Their second point of intersection J lies on the 7-point circle; (iii) The line IJ passes through K, the symmedian point of ABC; (iv) Triangles YZX and Z'X'Y' are similar; (v) Lines YX', ZY', XZ' are concurrent at Q, the Spieker centre of ABC; (vi) If U, V, W are the midpoints of YZ', ZX', XY' respectively then circle UVW has OI as a diameter and passes through J; (vii) Circles AYC, BZA, CXB pass through the Brocard point Ω and circles AY'C, BZ'A, CX'B pass through the Brocard point Ω'; (viii) Circle AYΩC and BZ'Ω'A meet at a point X'' lying on the 7-pt circle, circle BZΩA and CX'Ω'B meet at a point Y'' lying on the 7-pt circle and circle CXΩB and AY'Ω'C meet at a point Z'' also lying on the 7-pt circle; (ix) Circles YZZ', ZXX', XYY' have a common point S' lying on circle X'Y'Z'; (x) Circles X'Y'X, Y'Z'Y, Z'X'Z have a common point S lying on circle XYZ; (xi) Circles AYZ, BZX, CXY have a common point D lying on the circumcircle, circles AY'Z', BZ'X', CX'Y' have a common point D' lying on the circumcircle and circles AY''Z'', BZ''X'', CX''Y'' have a common point D'' lying on the circumcircle. (xii) Circles BX''C, CY''A, AZ''B all pass through O. The line of centres of circles XYZ and X'Y'Z' is, of course perpendicular to IJ through its midpoint, but the centres do not appear to be significant. Results (i) to (viii) are illustrated in Figure 1 and results (ix) to (xii) are illustrated in Figure 2. Results are proved in the following sections using areal-co-ordinates with ABC as triangle of reference. 2. The angle bisectors and perpendicular bisectors The equations of the angle bisectors AI, BI, CI are respectively cy = bz, az = cx, bx = ay. The equation of the perpendicular bisector of BC is the line OL, with equation (b4 – c4 – a2b2 + a2c2)x + a2(b2 + c2 – a2)(y – z) = 0.
2
(2.1)
The equations of the perpendicular bisectors of CA, AB may be obtained from Equation (2.1) by cyclic change of x, y, z and a, b, c,.
A
X
O
Z'' Y'
W
Y S'
M
N
J
X''
Q U
G S Z
V X'
Y''
I Z'
K
' D'
B
C
L
D
D'' F ig ure 2
3. The six points X, Y, Z, X', Y', Z'
3
The co-ordinates of these six points are as follows: Y(bc + c2 – a2, b2, bc), Z(ca, ca + a2 – b2, c2), X(a2, ab, ab + b2 – c2), X'(a2, ca + c2 – b2, ca), Y'(ab, b2, ab + a2 – c2), Z'(bc + b2 – a2, bc, c2). 4. The circles XYZ and X'Y'Z' and their common chord IJ The circle XYZ has equation b2cx2 + c2ay2 +a2bz2 – a(ca + a2 – c2)yz – b(ab + b2 – a2)zx – c(bc + c2 – b2)xy = 0.
(4.1)
It may now be checked that I(a, b, c) lies on this circle and also that it passes through the Brocard point Ω(1/b2, 1/c2, 1/a2). The circle X'Y'Z' has equation bc2x2 + ca2y2 + ab2z2 – a(ab + a2 – b2)yz – b(bc + b2 – c2)zx – c(ca + c2 – a2)xy = 0.
(4.2)
It may now be checked that I(a, b, c) lies on this circle and also that it passes through the Brocard point Ω'(1/c2, 1/a2, 1/b2). The second point of intersection of the two circles is J, which has co-ordinates x = a(a4 – a3(b + c) – a2bc + 2abc(b + c) – bc(b2 + c2)), y = b(b4 – b3(c + a) – b2ca + 2abc(c + a) – ca(c2 + a2)), z = c(c4 – c3(a + b) – c2ab + 2abc(a + b) – ab(a2 + b2).
(4.3)
It may now be checked that J lies on the 7-point circle with equation b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0.
(4.4)
The equation of IJ is bc(b – c)x + ca(c – a)y + ab(a – b)z = 0.
(4.5)
It is immediate that the line IJ passes through the symmedian point K(a2, b2, c2). 5. The similarity of triangles YZX and Z'X'Y' The circles YZX and Z'X'Y' intersect at I and J, and YZ', ZX', XY' all pass through I. It follows from Wood [1], that triangle Z'X'Y' is similar to YZX by means of a direct similarity composed of a rotation about J followed by an enlargement. 6. The concurrence of YX', ZY', XZ' at the Spieker centre Q The equation of YX' is 4
b(b – c)x + a(a – c)y – (a2 + b2 – ab – c2)z = 0.
(6.1)
The equations of ZY' and XZ' follow from Equation (6.1) by cyclic change of x, y, z and a, b, c. These three lines all pass through Q(b + c, c + a, a + b), which is the Spieker centre lying on line IG, where G is the centroid of ABC and IG = 2 GQ. 7. Points U, V, W and properties of circle UVW U is the midpoint of YZ' and so has co-ordinates U((b + c)2 – 2a2, b(b + c), c(b + c)). Similarly V has co-ordinates V(a(c + a), (c + a)2 – 2b2, c(c + a)) and W has co-ordinates W(a(a + b), b(a + b), (a + b)2 – 2c2). The equation of the circle UVW may now be obtained and it has equation bc(b + c)x2 ca(c + a)y2 + ab(a + b)z2 – a(2a2 – b2 – c2 + a(b + c))yz – b(2b2 – c2 – a2 + b(c + a))zx – c(2c2 – a2 – b2 + c(a + b))xy = 0. (7.1) It now follows that this circle passes through O and I and has centre the midpoint of OI and furthermore passes through J. 8. Circles YZZ', ZXX' and XYY' and the point S' lying on circle X'Y'Z' The equation of the circle YZZ' is b2c2x2 + c2(c(a + b) – a2)y2 + bc(a2 + b2 – ca)z2 + (a4 – a2(b2 + bc + 2c2) + ac2(c – b) + bc(b2 + c2))yz + b(a2(b + c) – ac2 – b2(b + c))zx + c3(a – b – c)xy = 0.
(8.1)
Circles ZXX' and XYY' have equations that may be written down from Equation (8.1) by cyclic change of x, y, z and a, b, c. It may now be verified that these circles have a common point S' with co-ordinates (x, y, z), where x = a(a3(b + c) – a2(b + c) – abc(b + 2c) + b2c(b – c)), y = – b(a2(b + c)2 – a(b3 – bc2 + c3) – b2c(b – c)), (8.2) 3 2 2 2 2 2 2 z = c(a b – a (b + bc + c ) + ac(c – 2b ) – bc (b – c)). It may now be checked that S' lies on circle X'Y'Z'. It follows from Result 2 in the Appendix that circles X'Y'Z, Y'Z'X, Z'X'Y have a common point S lying on circle XYZ. 9. The connection with the Brocard points Circle AYC has equation 5
c2y2 + (c2 – a2)yz – b2zx = 0.
(9.1)
It may now be checked that this circle passes through the Brocard point Ω (1/b2, 1/c2, 1/a2). Similarly circles BZA and CXB pass through Ω. These three circles are, of course, the circles that are often chosen to define Ω. Their appearance in the context of the angle bisectors is mildly surprising. Circle AY'C has equation a2y2 – b2zx + (a2 – c2)xy = 0. (9.2) 2 2 2 This circle passes through the other Brocard point Ω' (1/c , 1/a , 1/b ), as do the circles BZ'A and CX'B. The equation of the circle BZ'A is b2z2 + (b2 – a2)yz – c2xy = 0.
(9.3)
Circles AYΩC and BZ'Ω'A meet at X'' with co-ordinates (b2 + c2 – a2, b2, c2). Similarly circles BZΩA and CX'Ω'B meet at Y' with co-ordinates (a2, c2 + a2 – b2, c2) and circles CXΩB and AY'Ω'C meet at Z'' with co-ordinates (a2, b2 a2 + b2 – c2). Points X'', Y'', Z'' are the vertices of the second Brocard triangle and lie on the 7-point circle with Equation (4.4). 10.
The points D, D', D''
It has already been established that circles AYC, BZA, CXB pass through Ω, as does circle XYZ. It follows from Result 2 of the Appendix that circles AYZ, BZX, CXY meet at a point D on the circumcircle of ABC. Similarly circles AY'C, BZ'A, CX'B pass through Ω', as does circle X'Y'Z'. It follows again from Result 2 of the Appendix that circles AY'Z', BZ'X', CX'Y' meet at a point D' on the circumcircle of ABC. The equation of the circle CX''Y'' is b2c2(b2 – c2)x2 – c2a2(c2 – a2)y2 – a2(a4 – a2b2 – a2c2 + b4)yz – b2(a4 – a2b2 + b4 – b2c2)zx + c2(a2 – c2)(b2 – c2)xy = 0. (10.1) Circles AY''Z'' and BZ''X'' have similar equations that may be written down by cyclic change of x, y, z and a, b, c. These circles meet at the point D'' with co-ordinates (x, y, z), where x = a2(a2 – b2)(c2 – a2), y = b2(b2 – c2)(a2 – b2), (10.2) 2 2 2 2 2 z = c (c – a )(b – c ). It may now be checked that this point lies on the circumcircle of ABC.
6
The equation of circle AY''C is c2a2y2 + a2(b2 – a2)yz – b2(c2 + a2 – b2)zx + c2(b2 – c2) xy = 0.
(10.3)
Circles BZ''A and CX''B have similar equations to Equation (10.3) and may be written down by cyclic change of x, y, z and a, b, c. It may now be checked that all these circles pass through the circumcentre O. 11.
Another case with a similar configuration
It may be asked whether there are other examples of pairs of three lines whose intersections have such interesting properties. And the answer to this question is that the author knows of one other case. This is when one has the three altitudes AH, BH, CH and the three radii of the circumcircle AO, BO, CO. Points are now defined with X'= BH^CO, Y' = CH^AO, Z' = AH^BO, X = CH^BO, Y = AH^CO, Z = BH^AO. It then follows that circles XYZ, XY'Z', YZ'X', ZX'Y' all pass through the Brocard point Ω and the circles X'Y'Z', X'YZ, Y'ZX, Z'XY all pass through the Brocard point Ω'. It is also the case that circles XYZ and X'Y'Z' meet at H and again at a point J of the 7-point circle. See Figure 3.
7
A
Z' Y'
J X O
'
X'
H Y
Z
C
B
Figure 3
Proofs of all these results appear in Article 45 of Bradley [2].
Appendix 8
Some Observations on Pairs of Circles ABC and PQR having an intertwining property We say that circles ABC and PQR intertwine if circles ABR, BCP, CAQ pass through the same point S. Result 1
The intertwining property is symmetric
Thus, if circles BCP, CAQ, ABR have a common point S, then circles QRA, RPB, PQC have a common point D. Proof If S is the point of concurrence of circles BCP, CAQ and ABR, then invert with respect to S and the configuration becomes one in which (using stars for inverted points) P* lies on B*C*, Q* lies on C*A* and R* lies on A*B*. It then follows that circles Q*R*A*, R*P*B*, P*Q*C* are concurrent at the point D*, the Miquel point. The inverse image D of the point D* is now the point of concurrence of circles QRA, RPB and PQC.
Result 2
If in Result 1 the point S lies on circle PQR, then the point D lies on circle ABC
Proof See Result 1. If S now lies on circle PQR, after inverting with respect to S, the point D* is the point of intersection of the four circumcircles of the four triangles formed by the lines P*Q*R*, P*B*C*, Q*C*A*, R*A*B*. The inverse image D of the point D* now lies on circle ABC. See the Figure above. Result 3
The intertwining property is transitive
Thus if ABC and PQR intertwine and ABC and QRP also intertwine, then ABC and RPQ also intertwine. This result is not required in this article, but is added for the sake of completeness. This time, after inversion, the result is still not immediate, and the details are left to the reader. A proof appears in Article 19, by the Author [2] on the Internet, entitled ‘Circular Perspective’.
9
A
Q
Q* A*
P*
P
B*
S C*
R
C
B D* R* D
Acknowledgement I am grateful to David Monk [3] for pointing out the extension of Result 1 above, recorded as Result 2 and also for many useful exchanges about Section 11.
10
References 1. F.E. Wood, Amer. Math. Monthly 36:2 (1929) 67-73. 2. C.J.Bradley, Article 19 ,http://people.bath.ac.uk/masgcs/bradley.html 3. D. Monk, private communication.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
11
Three Triangles in Mutual Triple Perspective Christopher Bradley
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1. Introduction Given a triangle ABC and P a point not on its sides, the point P', the isotomic conjugate of P is selected and then the following points are constructed: L = AP^BP', M = BP^CP', N = CP^AP', L' = AP'^BP, M' = BP'^CP, N' = CP'^AP. It then turns out that triangles ABC, LMN, L'M'N' are in mutual triple perspective. Of the nine perspectors, three are at P, three are at P' one at W, one at W' and one at V. Here W is the perspector of triangles ABC and MNL, W' is the perspector of triangles ABC and M'N'L' and V the perspector of triangles LMN and L'M'N'. Of the nine perspectrices three coincide with DEF and one each with XYZ, X'Y'Z', RST, R'S'T', GHK and G'H'K'. These latter six perspectrices have a common point Q. W' turns out to be the isotomic conjugate of W. Also L and L', M and M', N and N' are isogonal conjugate pairs. The Figure 1
above displays this information. Points on the perspectrices are defined in later sections, when appropriate. The analysis is carried out using areal co-ordinates with ABC as triangle of reference with the help of DERIVE and the drawing is by CABRI. 2. The points P, P', L, L', M, M', N, N' Suppose P has co-ordinates (p, q, r), then P' has co-ordinates (qr, rp, pq). L = AP^BP' and since AP has equation qz = ry and BP' has equation rz = px, the co-ordinates of L are (r2, pq, rp). Similarly L' = AP'^BP and since AP' has equation qy = rz and BP has equation pz = rx, the coordinates of L' are (pq, r2, qr). Note that L'is the isotomic conjugate of L. The co-ordinates of M, N may be written down from those of L by cyclic change of x, y, z and p, q, r. Likewise those of M', N' may be written down by cyclic change from those of L'. Thus we have M(pq, p2, qr), N(rp, qr, q2), M'(pr, qr, p2) and N'(q2, pq, rp). Note that M', N' are the isogonal conjugates of M, N respectively. 3. The equation of the lines MN, NL, LM, M'N', N'L', L'M' From the co-ordinates of points in Section 2 we find the lines have equations: MN: q2(r2 – p2)x + pq(q2 – r2)y + rp(p2 – q2)z = 0; NL: pq(q2 – r2)x + r2(p2 – q2)y + qr(r2 – p2)z = 0; LM: rp(p2 – q2)x +qr(r2 – p2)y + p2(q2 – r2)z = 0; M'N': pq(r2 – p2)x + p2(q2 – r2)y + qr(p2 – q2)z = 0; N'L': rp(q2 – r2)x + qr(p2 – q2)y + q2(r2 – p2)z = 0; L'M': r2(p2 – q2)x + pq(r2 – p2)y + rp(q2 – r2)z = 0.
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
4. The points D, E, F, V, X, Y, Z, X', Y', Z', Q and the lines DEF, XYZ, X'Y'Z' It is now time to start studying the perspectives. First take triangles LMN and L'M'N'. The equation of LL' is rp(q2 – r2)x – qr(r2 – p2)y + (r4 – p2q2)z = 0. (4.1) The equations of MM' and NN' may be obtained from this by cyclic change of x, y, z and p, q, r. It may now be checked that these three lines all pass through the point V with co-ordinates (p(q2 + r2), q((r2 + p2), r(p2 + q2)). The lines MN and M'N meet at the point D with co-ordinates (p(q2 – r2), – q(r2 – p2), 0). The lines NL and N'L' meet at the point E with co-ordinates (0, – q(r2 – p2), r(p2 – q2)) and the lines LM and L'M' meet at the point F with co-ordinates (p(q2 – r2), 0, – r(p2 – q2)). The line DEF has equation x/p(q2 – r2) + y/q(r2 – p2) + z/r(p2 – q2)) = 0. (4.2)
2
Next we take the triangles ABC and LMN. They are clearly in perspective (by construction) at the point P. Now MN and BC meet at X(0, r(p2 – q2), – q(q2 – r2)), NL and CA meet at Y(– r(r2 – p2), 0, p(q2 – r2)) and LM meets AB at Z(q(r2 – p2), – p(p2 – q2), 0). The perspectrix XYZ has equation px/(r2 – p2) + qy/(p2 – q2) + rz/(q2 – r2) = 0. (4.3) Next we take the triangles ABC and X'Y'Z'. These are clearly in perspective (by construction) at the point P'. Now BC and M'N' meet at X' with co-ordinates (0, qr(p2 – q2), – p2(q2 – r2)), CA and N'L' meet at Y' with co-ordinates (– q2(r2 – p2), 0, rp(q2 – r2)) and AB and L'M' meet at Z' with co-ordinates (pq(r2 – p2), – r2(p2 – q2), 0). The perspectrix X'Y'Z' has equation r2px/(r2 – p2) + p2qy/(p2 – q2) + q2rz/(q2 – r2) = 0. (4.4) The perspectrices XYZ and X'Y'Z' meet at the point Q with co-ordinates (1/p(q2 – r2), 1/q(r2 – p2), 1/r(p2 – q2)). The point Q is significant as it lies on other perspectrices to be determined. 5. The perspectrices RST and R'S'T' We next consider triangles ABC and NLM which have perspector P'. LM meets BC at R with co-ordinates (0, p2(q2 – r2), – qr(r2 – p2). MN meets CA at S with co-ordinates (– rp(p2 – q2), 0, q2(r2 – p2)) and NL meets AB at T with co-ordinates (r2(p2 – q2), – pq(q2 – r2), 0). The perspectrix RST has equation pq2x/(p2 – q2) + qr2y/(q2 – r2) + rp2z/(r2 – p2) = 0. (5.1) It may now be checked that RST passes through Q. We next consider triangles ABC and N'M'L' which have perspector P. L'M' meets BC at R' with co-ordinates (0, r(q2 – r2), – q(r2 – p2)). M'N' meets CA at S' with co-ordinates (– r(p2 – q2), 0, p(r2 – p2)) and N'L' meets AB at T' with co-ordinates (q(p2 – q2), – p(q2 – r2), 0). The perspectrix R'S'T' has equation px/(p2 – q2) + qy/(q2 – r2) + rz/(r2 – p2) = 0. (5.2) It may now be checked that R'S'T' passes through Q. 6. The other two perspectors W and W' These arise from triangles ABC and MNL, and ABC and M'N'L' respectively and have coordinates W(r2p, p2q, q2r) and W'(q2p, r2q, p2r). It is interesting to note that W and W' are isotomic conjugates. The line WW' has equation 3
qr(p4 – q2r2)x + rp(q4 – r2p2)y + pq(r4 – p2q2)z = 0.
(6.1)
It is noteworthy that WW' passes through V, like other pairs of isotomic conjugates L, L' and M, M' and N, N'. As can be seen from the figure the perspectrices of these perspectives both coincide with DEF. 7. The final pair of perspectives These arise from the pair LMN and M'N'L' with perspector P' and the pair LMN and N'L'M' with perspector P. Sides MN and N'L' meet at the point G with co-ordinates (pq(r2p4 + p2q4 + q2r4 – 3p2q2r2), p4q4 + q4r4 + r4p4 – p4q2r2 – q4r2p2 – r4p2q2, qr(p4q2 + q4r2 + r4p2 – 3p2q2r2)). The coordinates of points H, K may be written down from those of G by cyclic change of x, y, z and p, q, r. Sides MN and L'M' meet at the point G' with co-ordinates (pqr(p4 + q4 + r4 – p2q2 – q2r2 – r2p2), r(r2p4 + p2q4 + q2r4 – 3p2q2r2), q(p4q2 + q4r2 + r4p2 – 3p2q2r2)). The co-ordinates of points H', K' may be written down from those of G' by cyclic change of x, y, z and p, q, r. The equations of lines GHK and G'H'K' are too complicated to record, but DERIVE has checked that both these perspectrices also pass through Q.
Flat 4 Terrill Court 12-14 Apsley Road BRISTOL BS8 2SP
4
When one Conic produces two more Christopher J Bradley
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1. Introduction In the Figure a triangle ABC is shown and O is a point not on its sides. Lines BO, CO are drawn and points M and N on these lines are marked. A conic ABCMN is now drawn meeting the line AO at L. Further points are defined as follows LM^BC = X, MN^CA = Y, NL^AB = Z, MN^AB = D, NL^BC = E, LM^CA = F, AL^BC = P, BM^CA = Q, CN^AB = R, AL^MN = U, BM^NL = V and CN^LM = W. The following results now hold: (i) A conic may be drawn through XYZDEF; (ii) A conic may be drawn through PQRUVW; (iii) Triangles XYZ and UVW are in perspective. It is, of course trivial that triangles PQR and UVW are also in perspective. Analysis
is carried out using projective co-ordinates with ABC as triangle of reference and O as unit point (1, 1, 1). 2. The conic ABCLMN The points L, M, by construction have co-ordinates of the form L(f, 1, 1) and M(1, g, 1). The conic through A, B, C, L, M has equation f(1 – g)yz + g(1 – f)zx + (fg – 1)xy = 0. (2.1) If the point N(1, 1, h) lies on the conic then h = (fg – 1)/(2fg – f – g).
(2.2)
Consequently we may take N to have co-ordinates N(2fg – f – g, 2fg – f – g, fg – 1). 3. Lines LM, MN, NL and points X, Y, Z, L, M, N The co-ordinates of L, M, N are given in Section 2 and from them the equations of the lines LM, MN, NL turn out to be LM: (1 – g)x + (1 – f)y + (fg – 1)z = 0; (3.1) MN: f(g – 1)x + (f – 1)y – (2fg – f – g)z = 0; (3.2) NL: (g – 1)x + g(f – 1)y – (2fg – f – g)z = 0. (3.3) Co-ordinates of points now follow easily and are: X = BC^LM (0, fg – 1, f – 1); Y = CA^MN (2fg – f – g, 0, f(g – 1); Z = AB^NL (g(f – 1), 1 – g, 0); D = AB^MN (f – 1, f(1 – g), 0); E = BC^NL (0, 2fg – f – g, g(f – 1)); F = CA^LM (fg – 1, 0, g – 1); It may now be checked that triangles XYZ and DEF are in perspective with vertex O. Some hard work (by DERIVE) shows the conic XYZDEF has equation f(g – 1)2x2 + g(f – 1)2y2 + (fg – 1)(2fg – f – g)z2 + (1 – f)(f(g2 + 2g – 1) – 2g)yz + (1 – g)(g(f2 + 2f – 1) – 2f)zx + (f – 1)(g – 1)(1 + fg)xy = 0. (3.4) The fact that a conic passes through these six points has an easier ‘pure’ proof. The triangles ABC and LMN are in perspective so three of the nine points of intersection lie on a line (of degree one in x, y, z), the Desargues’ axis of perspective. The nine points being the intersection of two triangles must create a degenerate cubic, so the remaining six points of intersection X, Y, Z, D, E, F must lie on a conic.
4. The points P, Q, R, U, V, W and the conic passing through these points The points P, Q, R obviously have co-ordinates P(0, 1, 1), Q(1, 0, 1), R(1, 1, 0). The remaining three points have co-ordinates as follows: U = MN^y = z (2f – 1, f, f); V = NL^z = x (g, 2g – 1, g); W = LM^x = y (fg – 1, fg – 1, f + g – 2). It may now be checked that these six points lie on the conic with equation f(1 – g)x2 + g(1 – f)y2 + (fg – 1)z2 +(1 – g)yz + (1 – f)zx + (2fg – f – g)xy = 0. Triangles PQR and UVW are in perspective with vertex O.
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(4.1)
Article 52 Three triangles in mutual Triple Reverse Perspective Christopher J Bradley
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A1 B2 C3 P A2 B3 C1 Q A3 B1 C2 R
AQ BP CR 2 AR BQ CP 3 AP BR CQ 1
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P1 Q2 R3 A P2 Q3 R1 B P3 Q1 R2 C
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1. Introduction The three triangles concerned are triangles ABC, PQR and 132 (all having anticlockwise labelling). The figure is interesting in the sense that P, Q, R are the perspectors of triangles ABC and 123, that 1, 2, 3 are the perspectors of triangles ABC and QPR and that A, B, C are the perspectors of triangles PQR and 123. (Apologies to those who feel that labels 2 and 3 ought to
1
U
have been interchanged, but that would have meant a rewriting of the analysis and that might well have introduced errors.) Note how the figure is drawn in order to provide total accuracy. First triangle ABC and points P and Q were chosen. Then lines AP, BP, CP, AQ, BQ, CQ were drawn. Now point 1 was defined as AP^CQ and point 2 was defined as AQ^BP. Then point 3 was defined as CP^BQ. Lines B1, C2, A3 are then found to be concurrent at the point R. The method of construction ensures the mutual triple reverse perspectives mentioned above. No further analysis is actually needed to establish this, but in view of intended applications it becomes necessary to set up the analysis and to have recorded the co-ordinates of all points and the equations of the nine perspectrices. Areal co-ordinates are used with ABC as triangle of reference. 2. Points P, Q, R, 1, 2, 3 We take P to have co-ordinates (p, q, r) and Q to have co-ordinates (l, m, n). Then lines AP, BP, CP have equations ry = qz, rx = pz and qx = py respectively. LinesAQ, BQ, CQ have equations ny = mz, nx = lz and mx = ly respectively. AP meets CQ at point 1(lq, mq, mr). BP meets AQ at point 2(np, mr, nr). CP meets BQ at point 3(lp, lq, np). B1 has equation mrx = lqz, A3 has equation npy = lqz and C2 has equation mrx = npy. R is the point of concurrence of these lines and has co-ordinates (lqnp, lqmr, mrnp). 3. Lines 12, 23, 31 and Desargues’ axes LMN, DEF, UVW Line 12 has equation mr(mr – nq)x + nr(lq – mp)y + mq(np – lr) z = 0.
(3.1)
Line 23 has equation nr(lq – mp)x – pn(lr – np)y – pl(nq – mr)z = 0.
(3.2)
Line 31 has equation mq(lr – np)x – lp(mr – nq)y – lq(lq – mp)z = 0.
(3.3)
Now A1, B2, C3 meet at P and the Deasargues’ axis of this perspective is LMN, where L = BC^23, M = CA^31, N = AB^12. L has co-ordinates (0, l(mr – nq), n(lr – np)). M has co-
2
ordinates (l(mp – lq), 0, m(np – lr)). N has co-ordinates (n(lq – mp), m(nq – mr), 0). The axis LMN has equation mx/(lq – mp) + ny/(mr – nq) + lz/(np – lr) = 0. (3.4) Triangles ABC and 231 are in perspective with vertex Q and the Desargues’ axis of perspective is DEF, where D = BC^31, E = CA^12, F = AB^23. D has co-ordinates (0, q(lq – mp), p(nq – mr)). E has co-ordinates (q(lr – np), 0, r(mr – nq)). F has co-ordinates (p(lr – np), r(lq – mp), 0). The axis DEF has equation rx/(lr – np) + py/(mp – lq) + qz/(nq – mr) = 0. (3.5) Triangles ABC and 312 are in perspective with vertex R and the Desargues’ axis of perspective is UVW, where U = BC^12, V = CA^23, W = AB^31. U has co-ordinates (0, mq(lr – np), nr(lq – mp)). V has co-ordinates (lp(mr – nq), 0, nr(mp – lq)). W has co-ordinates (lp(mr – nq), mq(lr – np), 0). The axis UVW has equation x/(lp(mr – nq)) + y/(mq(np – lr)) + z/(nr(lq – mp)) = 0. (3.6) 4. Lines PQ, QR, RP and axes D'E'F', L'M'N', U'V'W' The line PQ has equation (nq – mr)x + (lr – np)y + (mp – lq)z = 0.
(4.1)
The line QR has equation mnr(lq – mp)x + lnp(mr – nq)y + lmq(np – lr)z = 0.
(4.2)
The line RP has equation mqr(lr – np)x +pnr(mp – lq)y + plq(nq – mr)z = 0.
(4.3)
Triangles ABC and QPR are in perspective with vertex 2 and the Desargues’ axis of perspective is D'E'F' where D' = BC^PR, E' = CA^RQ, F' = AB^QP. D' has co-ordinates (0, lq(nq – mr), nr(lq – mp)). E' has co-ordinates (lq(lr – np), 0, nr(lq – mp)). F' has co-ordinates (lr – np, mr – nq, 0). The axis D'E'F' has equation nrx/(lr – np) + nry/(nq – mr) + lqz/(mp – lq) = 0. (4.4) Triangles ABC and RQP are in perspective vertex 3 and the Desargues’ axis of perspective is L'M'N', where L' = BC^QP, M' = CA^ PR, N' = AB^RQ. L' has co-ordinates (0, lq – mp, lr – np). M' has co-ordinates (lp(mr – nq), 0, mr(lr – np)). N' has co-ordinates (lp(nq – mr), mr(lq – mp), 0). The axis L'M'N' has equation mrx/(mr – nq) + lpy/(lq – mp) + lpz/(np – lr) = 0. (4.5)
3
Triangles ABC and PRQ are in perspective with vertex 1 and the Desargues’ axis of perspective is U'V'W', where U' = BC^RQ, V' = CA^QP, W' = AB^PR. U' has co-ordinates (0, mq(lr – np), np(mr – nq)). V' has co-ordinates (lq – mp, 0, nq – mr). W' has co-ordinates (np(lq – mp), mq(lr – np), 0). The axis U'V'W' has equation mqx/(lq – mp) + npy/(np – lr) + mqz/(mr – nq) = 0. (4.6) 5. Axes XYZ, X'Y'Z', X''Y''Z'' Triangles PQR and 123 are in perspective with vertex A and the Desargues’ axis of perspective is XYZ, where X = QR^23, Y = RP^31, Z = PQ^12. X has co-ordinates (lp(lmr2 – n2pq), lmr(lqr + mpr – 2npq), npr(2lmr – lnq – mnp)). Y has co-ordinates (lp(m2pr – lnq2), lmq(2mpr – npq – lqr), mpr(lmr + mnp – 2lnq)). Z has co-ordinates (l2qr – mnp2, m(2lqr – mpr – npq), r(lmr + lnq – 2mnp)). The axis XYZ has equation mr((l2qr(mr + nq) + lp(m2r2 – 6mnqr + n2q2) + mnp2(mr + nq))x – pr(l2(m2r2 – mnqr + n2q2) – lmnp(mr + nq) + m2n2p2)y – lm(l2q2r2 – lpqr(mr + nq) + p2(m2r2 – mnqr + n2q2))z = 0. (5.1) Triangles PQR and 231 are in perspective with vertex B and the Desargues’ axis of perspective is X'Y'Z', where X' = PQ^23, Y'= QR^31, Z' = RP^12. X' has co-ordinates (p(lmr + mnp – 2lnq), m2pr – lnq2, n(2mpr – npq – lqr)). Y' has co-ordinates (lpq(lmr + lnq – 2mnp), mq(l2qr – mnp2), mnp(2lqr – mpr – npq)). Z' has co-ordinates (npq(2lmr – lnq – mnp), mq(lmr2 – n2pq), mnr(lqr + mpr – 2npq)). The axis X'Y'Z' has equation mn(l2q2r2 – lpqr(mr + nq) + p2(m2r2 – mnqr + n2q2))x – pn(l2qr(mr + nq) + lp(m2r2 – 6mnqr + n2q2))y+ pq (l2(m2r2 – mnqr + n2q2) – lmnp(mr + nq) + m2n2p2)z = 0. (5.2) Triangles PQR and 312 are in perspective with vertex C and the Desargues’ axis of perspective is X''Y''Z'' where X'' = PR^32, Y'' = PQ^31, Z''= QR^12. X'' has co-ordinates (lnp(2lqr – pmr – pnq), lqr(lmr + lnq – 2mnp), nr(l2qr – mnp2)). Y'' has co-ordinates (l(lqr + pmr – 2npq), q(2lmr – lnq – mnp, lmr2 – n2pq). Z'' ha co-ordinates (lnq(lqr + npq – 2mpr), mqr(2lnq – lmr – mnp), nr(lnq2 – m2pr). The axis X''Y''Z'' has equation qr(l2(m2r2 – mnqr + n2q2) – lmnp(mr + nq) + m2n2p2)x + ln(l2q2r2 – lpqr(mr + nq) + p2(m2r2 – 6mnqr + n2q2))y – lq(l2qr(mr + nq) + lp(m2r2 – 6mnqr + n2q2) + mnp2(mr + nq))z = 0. (5.3)
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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ARTICLE 53 Hagge Circles touching at H Christopher Bradley
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1. Introduction A construction is described that provides circumcircles to a pair of triangles, circles which touch at the orthocentre of ABC, and are hence touching Hagge circles. Analysis is given proving the 1
results by means of areal co-ordinates, with (u, v, w) for H and ABC as reference triangle. The triangles are enlargements /reductions of ABC, the centres of enlargement being vertices of perspective. Rather curiously for each triangle and its circumcircle the centre of the circle, the vertex of perspective and the nine-point centre N are collinear. This is, of course, related to Peiser’s [1] result for Hagge circles. In the Figure above, triangle 123 is in perspective with ABC through point 4, circle 123 having centre 0 and triangle PQR is in perspective with ABC through point S, circle PQR having centre O and touching circle 123 at the point 3 (which coincides with H). The construction is as follows: Take 3 to be the orthocentre H, and draw a line parallel to BC through 3 and choose a point 2 on this line. Point 1 is now fixed, being the intersection of lines through 2 and 3 parallel to AB and AC respectively. Points P, Q, R are now found so that P = B3^C2, Q = C1^A3, R = A2^B1. Results now are : (i) AP, BQ, CR are concurrent at a point S;(ii) A1, B2, C3 are concurrent at a point 4; (iii) circles 123 and PQR touch at 3; (iv) N, S, O are collinear; and (v) N, 0, 4 are collinear. 2. Preliminary analysis Point 3 has co-ordinates (u, v, w) and as 3 is the orthocentre we know u, v, w, are proportional to 1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2) respectively. To get the centre of enlargement we write 4 = k3 + (1 – k)C. (2.1) Writing t = (1 – k)/k and dividing Equation (2.1) by k the co-ordinates of 4 may be written as (u, v, w + t). Since 21 is parallel to BA and 31 is parallel to CA, the co-ordinates of 2 are (u v – t, z = w + t) and those of 1 are (u – t, v, w + t). Using these co-ordinates we find the equations of lines needed to find P, Q, R are as follows: A2: z(v – t) = y(w + t); (2.2) B1: x(w + t) = z(u – t); (2.3) B3: xw = zu; (2.4) C2: x(v – t) = yu; (2.5) A3: zv = yw; (2.6) C1: y(u – t) = xv. (2.7) The co-ordinates of P = B3^C2 are therefore (u, v – t, w), the co-ordinates of Q = B3^C2 are therefore (u – t, v, w) and the co-ordinates of R = A2^B1 are (u – t, v – t, w + t). 3. The circles PQR and 123
2
The equation of any circle, using u, v, w as the co-ordinates of H, is u(v + w)yz + v(w + u)zx + w(u + v)xy + (px + qy + rz)(x + y + z) = 0, where p, q, r are to be determined from the co-ordinates of three points on the circle.
(3.1)
Using the co-ordinates of P, Q, R we find the equation of the circle PQR is vw(v + w)x2 + wu(w + u)y2 – uv(2t – u – v)z2 – u(t(v – w) + 2vw)yz – v(t(u – w) + 2uw)zx + w(t(u + v) – 2uv)xy = 0. (3.2) Repeating the exercise using the co-ordinates of 1, 2, 3 we find the equation of the circle 123 is vw(t + u + v + w)(v + w)x2 + wu(t + u + v + w)(w + u)y2 – uv(tu + tv + 2tw – u2 – v2 – 2uv – uw – vw)z2 – u(tuv – tuw + tv2 + 2tvw – tw2+ 2vw(u + v + w))yz – v(t(u2 + uv + 2wu) – twv – tw2) + 2uw(u + v + w))zx + w(t(u2 + uw + v2 + vw) – 2uv(u + v + w))xy = 0. (3.3) The midpoint of 3R has co-ordinates 1/(2(u + v + w)(u + v + w – t))(2u2 + 2u(v + w – t) – t(v + w), u(2v – t) + 2v2 + 2v(w – t) – tw, u(2w + t) + v(2w + t) + 2w2). (3.4) It may now be checked that the co-ordinates of O, the centre of circle PQR are the same as those of the midpoint of 3R, given by Equation (3.4). The centre 0 of circle 123 may now be calculated and its co-ordinates are (2u(u + v + w) – t(v + w), 2v(u + v + w) – t(w + u), t((u + v + 2w) + 2w(u + v + w))). (3.5) It now follows that 0 lies on the line 3R. The conclusion from this is that circles 123 and PQR touch at 3 and that 3R is the diameter of circle PQR. 4. Conclusions It is easy to show that the point S = AP^BP^CP has co-ordinates (u – t, v – t, w). The nine-point centre N has co-ordinates (2u + v + w, u + 2v + w, u + v + 2w). And the vanishing of certain determinants establishes that 40N and OSN are straight lines. Reference 1. A.M. Peiser, The Hagge circle of a triangle, Amer. Math. Monthly, 49 (1942) 524-527.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article 54 Four Triangle Conics Christopher J Bradley
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1. Introduction
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A point P is taken in the plane of a triangle ABC and lines through P are drawn parallel to the sides of a triangle, creating six points on the sides of the triangle (as in the construction of the triplicate ratio circle when P is the symmedian point). These six points are found to lie on a conic S. Another 15 points may be constructed, as described below, from points of intersection of the lines joining these six points. 3 of these points lie on the polar of P with respect to S, and the remaining 12 lie 6 by 6 on two other conics. When P is the symmedian point, these two conics do not appear (as far as I am aware) in Brocard geometry and so do not appear to be related in any way to the 7-point circle or the Brocard ellipse. The Figure above shows these results, which will be clearer when the points in the figure have been carefully defined. 2. The six points L1, L2, M1, M2, N1, N2 on the sides of ABC The equation of the line parallel to BC through P is (q + r)x = p(y + z), the equation of the line through P parallel to CA is (r + p)y = q(z + x) and the equation of the line through P parallel to AB is (p + q)z = r(x + y). From these equations we may find the co-ordinates of the six points on the sides of the triangle. These are L1: (0, p + q, r); L2 (0, q, r + p); M1 (p, 0, q + r); M2 (p + q, 0, r); N1(p + r, q, 0); N2( p, q + r, 0). 3. The conic through the 6 points By substituting the co-ordinates of five of these points into the general equation of a conic and solving for the ratios of the six coefficients we find the equation of the conic through the five points is qr(q + r)x2 + rp(r + p)y2 + pq(p + q)z2 – p(p2 + pq + rp + 2qr)yz – q(q2 + rp+ pq + 2rp)zx – r(r2 + pq + qr + 2pq)xy = 0. (3.1) As a check it was then shown by substitution that the sixth point satisfied Equation (3.1). 4. Defining the points R, S, T, U, V, W and determining the conic through these six points The equations of the following six lines are: M1N1: q(q + r)x = (r + p)(q + r)y + pqz ; N1L1: r(r +p)y = (p + q)(r + p)z + qrx ; L1M1: p(p + q)z = (p + q)(q + r))x + rpy; M2N2: r(q + r)x = rpy + (p + q)(q + r)z; N2L2: p(r + p)y = pqz + (q + r)(r + p)x; L2M2: q(p + q)z = qrx + (r + p)(p + q)y; The six points are: R = L2N2^L1M1 (p2, (p + q)(q + r), (q + r)(r + p); S = L2M2^L1M1 (p2(p + q), q2(p + q), p(p + q)(q + r) + q2r); 2
T = M1N1^L2M2 ((p + q)(r + p), q2, (r + p)(q + r)); U = M1N1^M2N2 (q(q + r)(r + p) + r2p, q2(q + r), r2(q + r)); V = N1L1^M2N2 ((p + q)(r + p), (p + q)(q + r), r2); W = N1L1^N2L2 (p2(r + p), r(r + p)(p + q) + p2q, r2(r + p)). Using the same method as in Section 3 we find the equation of the conic RSTUVW is (q + r)(p(q + r) + q2 + r2)x2 + (r + p)(q(r + p) + r2 + p2)y2 + (p + q)(r(p + q) + p2 + q2)z2 + (p3 – p2(q + r) – 2p(q2 + qr + r2) – 2qr(q + r))yz + (q3 – q2(r + p) – 2q(r2 + rp + p2) – 2rp(r + p))zx + (r3 – r2(p + q) – 2r(p2 + pq + q2) – 2pq(p + q))xy = 0. (4.1) 5. The collinear points L, M, N and the point Q The points L, M, N are defined as follows: L = N1L1^L2M2 (– p(p + q)(r + p), qr(p + q), qr(r + p)); M = L1M1^M2N2 (rp(p + q), – q(q + r)(p + q), rp(q + r)); N = M1N1^ N2L2 (pq(r + p), pq(q + r), – r(q + r)(r + p)). It may now be checked that these three points lie on the line with equation qr(q + r)x + rp(r + p)y + pq(p + q)z = 0.
(5.1)
AL, BM, CN evidently concur at the point Q (1/(q + r), 1/(r + p), 1/(p + q)). 6. The points D, E, F, I, J, K and the conic passing through these points First three lines L2N1, M1N2, M2L1. Their equations are (in that order): (r + p)y = q(z + x); (q + r)x = p(y + z); (p + q)z = r(x + y). The six points have co-ordinates: D = L2N1^L1M1 (p2, q(p + q), pq + qr + rp); E = M1N2^L2M2 (p(p + q), q2, pq + qr + rp); F = M2L1^M1N1 (pq + qr + rp, q2, r(q + r)); I = N1L2^M2N2 (pq + qr + rp, q(q + r), r2); J = M1N2^N1L1 (p(r + p), pq + qr + rp, r2); K = L1M2^N2L2 (p2, pq + qr + rp, r(r + p)). A conic passes through these six points and has equation (p2(q2 + r2)(q + r)2 + pqr(q3 + q2r + qr2 + r3) – q3r3)x2 + ... + ... + (p4(q2 + qr + r2) – p3(q3 + r3) – p2qr(2q2 + 3qr + 2r2) – 3pq2r2(q + r) – 2q3r3)yz + ... + ....
7. The 7-point conic
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(6.1)
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In Figure 2 we show the construction of the (triplicate ratio) conic and the (7-point) conic when lines are drawn parallel to the sides of ABC through an arbitrary point P (rather than the symmedian point K). As you would expect these conics have the same centre, and PXQ in the Figure 2 is a diameter. W and W' are the analogues of the Brocard points and LMN the vertices of the first Brocard triangle. (In the figure D replaces L1 etc.) A final interesting fact is that if P is chosen to be the circumcentre O, then K lies on the 7-point conic, so that OK is a diameter in both cases, and X in Figure 2 is then the same as the centre of the 7-point circle. Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP 4
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Article 55 Perpendiculars to a Triangle’s Sides through its Vertices Christopher J Bradley
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1. Introduction
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The six points M1, N1, N2, L2, L3, M3 may be defined in three equivalent ways. First they are the points where the perpendiculars to the sides through the vertices of triangle ABC meet the opposite sides. For example L2 is the point on CA where the perpendicular to BC at B meets CA. Secondly, if LMN is the triangle formed by the tangents to the circumcircle of triangle ABC and XYZ is the triangle that has LMN as its medial triangle, then the six points are the points on the sides of ABC where the (non-corresponding) sides of XYZ meet the sides of ABC. For example L2 is the point common to XY and CA. Thirdly if one constructs the three circles through B, C and C, A and A, B that cut the circumcircle orthogonally, the six points are the intersections of these circles with the sides of ABC. For example, circle L2ABM1 is orthogonal to the circumcircle of ABC and cuts CA at L2 and BC at M1. In this article these equivalences are established and it is also shown that the six points lie on a conic. It is also shown that XA, YB, ZC are concurrent at a point R and that XL, YM, ZN are concurrent at a point Q. Work is carried out using areal co-ordinates with ABC as triangle of reference. 2. The triangle LMN and the lines through its vertices parallel to its opposite sides The circumcircle S of triangle ABC has equation a2yz + b2zx + c2xy = 0.
(2.1)
It follows that the tangents to S at A, B, C are respectively b2z + c2y = 0, c2x + a2z = 0 and a2y + b2x = 0. The point L is the intersection of the tangents at B and C and so has co-ordinates (– a2, b2, c2). Similarly M has co-ordinates (a2, – b2, c2) and N has co-ordinates (a2, b2, – c2). The equation of the line through L parallel to MN (the tangent at A) is 2b2c2x + c2(a2 + b2 – c2)y + b2(c2 + a2 – b2)z = 0.
(2.2)
The equations of the lines through M parallel to NL and through N parallel to LM may now be written down from Equation (2.2) by cyclic change of x, y, z and a, b, c. 3. The points X, Y, Z and the six points M3, N2, N1, L3, L2, M1 The line through L parallel to MN and the line through M parallel to NL meet at the point Z with co-ordinates (x, y, z), where x = – a2(a4 + 2a2(b2 – c2) – (b2 – c2)(3b2 + c2)), y = b2(3a4 – 2a2(b2 + c2) – (b2 – c2)2), (3.1) 2 4 2 2 2 2 2 2 z = – c (a – 2a (b + c ) + (b – c ) ). The co-ordinates of points X and Y may now be written down from Equation (3.1) by cyclic change of x, y, z and a, b, c.
2
We define the six key points as the intersections of the sides of triangle XYZ with the (noncorresponding) sides of ABC. Thus YZ meets AB at M3 and CA at N2. The co-ordinates of M3 are therefore (c2 – a2 – b2, 2b2, 0) and the co-ordinates of N2 are (b2 – c2 – a2, 0, 2c2). In similar fashion we may work out the co-ordinates of the other key points as N1(0, a2 – b2 – c2, 2c2), L3(2a2, c2 – a2 – b2, 0), L2(2a2, 0, b2 – c2 – a2), M1(0, 2b2, a2 – b2 – c2). 4. The circles M3BCN2, N1CAL3 and L2ABM1 and the conic through the six key points The equation of circle M3BCN2 is 2b2c2x2 – a2(b2 + c2 – a2)yz + b2(c2 + a2 – b2)zx + c2(a2 + b2 – c2)xy = 0.
(4.1)
The equations of circles N1CAL3 and L2ABM1 may now be written down from Equation (4.1) by cyclic change of x, y, z and a, b, c. The equation of the conic M3N2N1L3L2M1 is 2x2/a2 + ... + ... + (a4 + b4 + c4 + 6b2c2 – 2c2a2 – 2a2b2)/(b2c2(b2 + c2 – a2))yz + ... + ... = 0.
(4.2)
It is not surprising that the six key points lie on a conic as they are defined in such a consistent fashion. 5. AX, BY, CZ meet at a point R and XL, YM, ZN meet at a point Q These are not difficult to check. The x co-ordinate of R is x = – a2(a4 + b4 – 3c4 + 2b2c2 + 2c2a2 – 2a2b2)(a4 + c4 – 3b4 + 2b2c2 – 2c2a2 + 2a2b2). The y and z co-ordinates follow by cyclic change of a, b, c.
(5.1)
The x co-ordinate of Q is x = – a2(3a4 – b4 – c4 +2b2c2 – 2c2a2 – 2a2b2). The y and z co-ordinates follow by cyclic change of a, b, c.
(5.2)
6. The points D, E, F and their significance The line CO meets the circumcircle again at the point F with co-ordinates (x, y, z), where x = 2a2(b2 + c2 – a2), y = 2b2(c2 + a2 – b2), z = – (b2 + c2 – a2)(c2 + a2 – b2). The co-ordinates of D, E may now be written down by cyclic change of x, y, z and a, b, c. 3
(6.1)
The equation of the line BF is (c2 + a2 – b2)(b2 + c2 – a2)x + 2a2(b2 + c2 – a2)z = 0.
(6.2)
It may now be checked that the point L2 lies on this line. Note that since CO is a diameter of S, the line BF is perpendicular to BC through B. This establishes the third property of the six key points; that they lie on the six lines through the vertices perpendicular to the sides. 7. Three final points It may now be checked that (i) The midpoints of YZ, ZX, XY are respectively the points L, M, N; (ii) L is also the midpoint of M3N2 etc; (iii)The tangent to circle M3BCN2 at C passes through O, the centre of S, and hence each of the three circles is orthogonal to S.
Flat 4 Terrill Couret 12-14 Apsley Road BRISTOL BS8 2SP
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Article 56 The Cevian point Conic Christopher Bradley
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1. Introduction In triangle ABC let P be a Cevian point with D, E, F the feet of the Cevians on BC, CA, AB respectively. Points L, M, N lie on AD, BE, CF respectively and are such that P is the midpoint of AL, BM, CN. It is found that the points A, B, C, L, M, N lie on a conic which we call the Cevian point conic of P. (When P lies at the centroid G this conic is the Steiner conic.) The tangents at A, B, C to the Cevian point conic form a triangle UVW. The midpoints of the sides BC, CA, AB are denoted by X, Y, Z respectively. The following results now hold: (i) DU, EV, FW are concurrent at a point Q; (ii) LU, MV, NW are concurrent at a point R; (iii) P, Q, R are 1
collinear; (iv) AU, BV, CW are concurrent at a point S; (iv) LX, MY, NZ are concurrent at a point T. 2. Co-ordinates of L, M, N and the equation of the Cevian point Conic Since AP = PL, BP = PM, CP = PN and P has co-ordinates (l, m, n), say, the co-ordinates of L, M, N are L(2l – 1, 2m, 2n), M(2l, 2m – 1, 2n), N(2l, 2m, 2n – 1). Note that we take l + m + n = 1. Then it is easily checked that the equation of the Cevian point conic is l(1 – 2l)yz + m(1 – 2m)zx + n(1 – 2n)xy = 0. (2.1) 3. Points U, V, W, D, E, F, X, Y, Z The equation of the tangent at A to the Cevian point conic is n(1 – 2n)y + m(1 – 2m)z = 0. The tangents at B and C may be written down by cyclic change of x, y, z and l, m, n. The tangents at B and C meet at U(l(1 – 2l), m(2m – 1), n(2n – 1)).Similarly the points V and W have co-ordinates V(l(2l – 1), m(1 – 2m), n(2n – 1)) and W(l(2l – 1), m(2m – 1), n(1 – 2n)). Clearly we have D(0, m, n), E(l, 0, n), F(l, m, 0), X(0, 1, 1), Y(1, 0, 1), Z(1, 1, 0). 4. DU, EV, FW are concurrent at Q The equation of DU is 2mn(n – m)x + l(1 – 2l)(ny – mz) = 0.
(4.1)
The equations of EV, FW follow by cyclic change of x, y, z and l, m, n. It then follows that the point Q with co-ordinates (x, y, z), where x = l(1 – 2l)(2l – 2m – 2n + 1), y = m(1 – 2m)(2m – 2n – 2l + 1), z = n(1 – 2n)(2n – 2l – 2m, + 1),
(4.2)
lies on all three lines. 5. LU, MV, NW are concurrent at R The equation of LU is 4mn(n – m)x + n(1 – 2l)(2n + 2l – 1)y – m(1 – 2l)(2l + 2m – 1)z = 0. The equations of MV, NW follow by cyclic change of x, y, z and l, m, n. 2
(5.1)
It then follows that the point R with co-ordinates (x, y, z), where x = l(1 – 2l)(2l + 2m + 2n – 8mn – 1), y = m(1 – 2m)(2l + 2m + 2n – 8nl – 1), z = n(1 – 2n)(2l + 2m + 2mn – 8lm – 1),
(5.2)
lies on all three lines. It may now be checked that the three points P, Q, R are collinear. 6. AU, BV, CW are concurrent at S The equation of AU is n(1 – 2n)y = m(1 – 2m)z.
(6.1)
The equations of BV, CW follow by cyclic change of x, y, z and l, m, n. It then follows that the point S with co-ordinates (l(2l – 1), m(2m – 1), n(2n – 1)) lies on all three lines. Note that when P is the symmedian point, then S is the circumcentre. 7. LX, MY, NZ are concurrent at T The equation of LX is 2(n – m)x + (2l – 1)(y – z) = 0.
(7.1)
The equations of MY, NZ follow by cyclic change of x, y, z and l, m, n. It then follows that the point T with co-ordinates (2l – 1, 2m – 1, 2n – 1) lies on all three lines. Below is the figure when P is the orthocentre, showing that even for an internal point, sometimes the Cevian point conic is a hyperbola. Note that when P is the symmedian point, T is the isotomic conjugate of the orthocentre.
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Article 57 Perpendiculars to the Cevians at the Cevian Point Christopher Bradley
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1. Introduction If ABC is a triangle and P a point not on the sides or extensions of the sides, then the lines AP, BP, CP are called the Cevians of P. If now the perpendicular to AP at P is drawn and it meets BC
at L, and M and N are similarly defined, it turns out that L, M, N are collinear. In this note a proof is given using areal co-ordinates with ABC as triangle of reference. 2. The displacements AP and PL Let the co-ordinates of P be (l, m, n), where l + m + n = 1, and those of L be (0, u, 1 – u). Then the displacement AP is (f, g, h) = (– m – n, m, n) and the displacement PL is (p, q, r) = (m + n – 1, u – m, 1 – u – n). The condition for these displacements to be perpendicular is, see Bradley [1], a2(gr + qh) + b2(ph + fr) + c2(fq + pg) = 0.
(2.1)
This provides the value for u, which is {a2m(2n – 1) – b2(m(2n – 1) + 2n(n – 1)) – c2m(2m + 2n – 1)}/{(b2 – c2)(m + n) – a2(m – n)} (2.2) If M has co-ordinates (1 – v, 0, v) and N has co-ordinates (w, 1 – w, 0) then v and w may be written down from equation (2.2) by cyclic change of a, b, c and l, m, n. 3. The collinearity of L, M, N L, M, N are collinear if and only if uv + vw + wu – u – v – w + 1 = 0.
(3.1)
Substitution of the values of u, v, w into the left hand side of equation (3.1) shows this to be true, since there is a factor of (l + m + n – 1) = 0.
Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, UK, (2007).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
Article 58 Additional results for the Miquel configuration Christopher J Bradley
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1. Introduction
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In the standard Miquel configuration ABC is a triangle with points L, M, N on BC, CA, AB respectively. Then circles C1= AMN, C2 = BNL, C3 = CLM have a point P in common. When LMN is a transversal then P lies on the circumcircle. The additional construction here is that the perpendiculars from L, M, N are drawn to the sides BC, CA, AB respectively. The one from L meets CA at M1 and AB at N1. The one from M meets BC at L2 and AB at N2. The one from N meets BC at L3 and CA at M3. It is a simple matter of angle-chasing to show that circles S1, S2, S3 may be drawn through M, N, N2, M3 and N, L, L3, N1 and L, M, M1, L2 respectively. The following results now hold: (i) Circles C1, S2, S3 have a point Q in common; (ii) Circles C2, S3, S1 have a point R in common; (iii) Circles C3, S1, S2 have a point S in common. It is also obvious from angular considerations that if the three perpendiculars form a triangle UVW then UVW is similar to ABC. When LMN is a transversal the following additional results hold: (iv) Points N2, M3, L3, N1, M1, L2 lie on a conic; (v) AU, BV, CW intersect at a point T, so that not only are UVW and ABC similar they are also in perspective with LMN the Desargues’ axis of perspective: (vi) T lies on the circumcircles of both ABC and UVW; (vii) If O1, O2, O3 are the centres of C1, C2, C3 respectively and O is the circumcentre of ABC, then it is known that O, O1, O2, O3, P are concyclic, but this circle also contains T and O1, O2, O3 lie on AT, BT, CT respectively, providing a Wood [1] configuration, establishing immediately the known similarity of triangles ABC and O1O2O3. We prove these results using areal co-ordinates with ABC as triangle of reference. 2. Co-ordinates of L, M, N In the general case we take L(0, u, 1 – u), M(1 – v, 0, v), N(w, 1 – w, 0). When LMN is the transversal with equation x/l + y/m + z/n = 0, this means that u = m/(m – n), v = n/(n – l), w = l/(l – m).
(2.1)
3. Co-ordinates of N1, L2, M3, M1, N2, L3 If (f, g, h) and (p, q, r) are two displacements at right-angles, it is known, see Bradley [2], that a2(gr + hq) + b2(fr + hp) + c2(fq + gp) = 0.
(3.1)
Suppose then that N1 has co-ordinates (s, 1 – s, 0) then the displacement BC = (0, – 1, 1) and the displacement LN1 = s, 1 – s – u, u – 1). Using these values for f, g, h, p, q, r we obtain s = 2a2(1 – u)/(c2 + a2 – b2).
(3.2)
It follows that N1 has co-ordinates (x, y, z), where x = 2a2(1 – u), y = 2a2u – a2 – b2 + c2, z = 0.
(3.3)
It follows by cyclic change of x, y, z and a, b, c and u, v, w that L2 has co-ordinates (x, y, z), where x = 0, y = 2b2(1 – v), z = 2b2v – b2 – c2 + a2. And again by cyclic change that M3 has co-ordinates (x, y, z), where 2
(3.4)
x = 2c2w – c2 – a2 + b2, y = 0, z = 2c2(1 – w).
(3.5)
If M1 has co-ordinates (1 – t, 0 , t) then the displacement LM1 is (1 – t, – u, t – 1 + u) and using Equation (3.1) we find t = {a2(1 – 2u) + b2 – c2}/(a2 + b2 – c2).
(3.6)
It follows that M1 has co-ordinates (x, y, z), where x = 2a2u, y = 0, z = a2(1 – 2u) + b2 – c2.
(3.7)
It now follows by cyclic change that the co-ordinates of N2 are (x, y, z), where x = b2(1 – 2v) + c2 – a2, y = 2b2v, z = 0.
(3.8)
And finally, L3 has co-ordinates (x, y, z), where x = 0, y = c2(1 – 2w) + a2 – b2, z = 2c2w.
(3.9).
4. The Miquel Point The Miquel point P is the common point of circles AMN, BNL, CLM. We first obtain the equation of circle AMN. Every circle in areal co-ordinates has an equation of the form a2yz + b2zx + c2xy + (px + qy + rz)(x + y + z) = 0.
(4.1)
The constants p, q, r are evaluated from equations found by inserting the co-ordinates of A, M, N in turn. The result is that circle AMN has equation c2wy2 + b2(1 – v)z2 – (a2 – b2(1 – v) + c2w)yz – b2vzx – c2(1 – w)xy = 0.
(4.2)
Circles BNL, CLM may now be written down by cyclic change of x, y, z and a, b, c and u, v, w. The Miquel point P is the common point of the three circles and has co-ordinates (x, y, z), where x = a2(a2u(u – 1) + b2(u – 1)(v – 1) + c2wu), y = b2(a2uv + b2v(v – 1) + c2(v – 1)(w – 1)),
(4.3)
z = c2(a2(w – 1)(u – 1) + b2vw + c2w(w – 1)). When L, M, N are collinear, then from Equations (2.1) the Miquel point P has co-ordinates x = {a2/(m – n)}, y = {b2/(n – l)}, z = {c2/(l – m)}.
(4.4)
It is easy to check that in this case P lies on the circumcircle of ABC. 5. The circles S1, S2, S3 The circle S1 is the circle passing through the points M, N, M3, N2. That a circle passes through these points follows immediately from the construction because of the right angles. Its equation is 3
2b2c2v(w – 1)x2 + c2w(a2 + b2(2v – 1) – c2)y2 + b2(1 – v)(a2 – b2 + c2(1 – 2w))z2 – (a4 + a2(b2(v – 2) – c2(w + 1)) + b4(1 – v) – b2c2(v(4w – 1) – 3w + 1) + c4w)yz – b2(a2v – b2v + c2(2(w – 1) – v(4w – 3)))zx + c2(a2(w – 1) + b2(2v(2w – 1) – w + 1) + c2(1 – w))xy = 0. (5.1) The equations of S2 and S3 may now be written down by cyclic change of x, y, z and a, b, c and u, v, w. It may now be verified that the point Q with co-ordinates (x, y, z) given next, lies on the three circles C1, S2, S3. x = a2(a4 + a2(b2(2v – 2) – 2c2w) + b4(1 – 2v) + b2c2(2v – 2w)) + c4(2w – 1))(a2u(u – 1) + b (u – 1)(v – 1) + c2wu), 2
y = – b2(a2(2u – 1) – b2 + c2)(a4uv – a2(b2v(u – v + 1) + c2(u(v + 2(w – 1)) + (v – 1)(w – 1))) + (1 – v)(b4v – b2c2(v – w + 1) + c4(1 – w))), (5.2) z = c2(a2(2u – 1) – b2 + c2)(a4(u – 1)(w – 1) – a2(b2(u(2v + w – 1) + v(w – 2) – w + 1) + c2(u – w – 1)(w – 1)) + w(b4v – b2c2(v – w + 1) + c4(1 – w))). Points R and S with co-ordinates provided by cyclic change lie similarly on circles C2, S3, S1 and C3, S1, S2 respectively. 6. The triangle UVW We now obtain the equations of the lies through L, M, N perpendicular to the sides that form the triangle UVW. The equation of the line LN1 is (a2(2u – 1) – b2 + c2)x + 2a2(u – 1)y + 2a2uz = 0.
(6.1)
The equations of the lines ML2 and NM3 may be written down from Equation (6.1) by cyclic change of x, y, z and a, b, c and u, v, w. LN1 and ML2 meet at the point W with co-ordinates (x, y, z), where x = – 2a2(a2u + b2(u + 2(v – 1)) – c2u), y = 2b2(a2(2u + v – 1) + (b2 – c2)(v – 1)),
(6.2)
z = a4(2u – 1) – 2a2(b2(u – v) + c2(u – 1)) + (c2 – b2)(b2(2v – 1) – c2). Points U and V have co-ordinates that may be written down from Equation (6.2) with cyclic change of x, y, z and a, b, c and u, v, w. It may be verified that W lies on C3 and U lies on C1 and V lies on C2. 7. The six points when L, M, N are collinear These may be obtained from Section 3 with the aid of Equation (2.1) the co-ordinates are: N1: (– 2a2n, a2(m + n) + (c2 – b2)(m – n), 0); L2: (0, – 2b2l, b2(n + l) + (a2 – c2)(n – l)); M3: (c2(l + m) + (b2 – a2)(l – m), 0, – 2c2m); 4
N2: (b2(n + l) + (c2 – a2)(l – n), – 2b2n, 0); L3: (0, c2(l + m) + (a2 – b2)(m – l), – 2c2l); M1: (– 2a2m, 0, a2(m + n) + (b2 – c2)(n – m)).
(7.1)
It may now be shown that these six points lie on the conic with equation 2b2c2lmn(a2(m + n) + (b2 – c2)(n – m))x2 + … + … + a2mn(a4(l – m)(l – n) – 2a2l(b2(l – m) + c2(l – n)) + b4(l – m)(l + n) + 2b2c2(3l2 + mn) + c4(l + m)(l – n))yz +… +… = 0. (7.2) 8. AU, BV, CW meet at T which lies on circles ABC and UVW & P also lies on circle UVW When L, M, N are collinear then, from Equations (6.2) and (2.1) the co-ordinates of W are (x, y, z), where x = – 2a2(a2m(l – n) + b2(l(m – 2n) + mn) + c2m(n – l)), y = 2b2(a2(l(m + n) – 2mn) + l(b2 – c2)(n – m)),
(8.1)
z = a4(l – n)(m + n) – 2a2(b2(lm – n2) + c2n(l – n)) + (b2 – c2)(m – n)(b2(l + n) + c2(l – n)). The co-ordinates of U and V now follow by cyclic change of x, y, z and a, b, c and l, m, n. It may now be shown that AU, BV, CW are concurrent at the point T with co-ordinates (x, y, z), where x = – a2/(a2(l(m + n) – 2mn) + l(b2 – c2)(n – m)), y = b2/(a2m(l – n) – b2(l(m – 2n) + mn) + c2m(n – l)),
(8.2)
z = c2/(a2n(l – m) + b2n(m – l) + c2(l(2m – n) – mn)). It may now be checked that T lies on the circumcircle of ABC. The circle UVW may now be obtained and has equation 2b2c2(m – n)(a2(lm + nl – 2mn) + (b2 – c2)l(n – m))x2 + … + … – (a4(l – m)(l – n)(m – n) – 2a2(b2(l – m)(l(m – 2n) + n2) + c2(l – n)(l(2m – n) – m2)) + b4(l – m)(l(m – 3n) + n(m + n)) + 2b2c2l(m – n)(2l – m – n) + c4(l – n)(l(3m – n) – m(m + n)))yz – … – … = 0. (8.3) It may now be verified that T lies on circle ABC and that P also lies on circle UVW. The Figure illustrating these results when L, M, N are collinear is shown below. It will be observed that triangle UVW is similar to triangle ABC by means of a rotation of 90 o and enlargement/reduction about P, since we have a Wood [1] configuration for similar triangles in perspective. It may also be shown that O1 lies on AT, O2 lies on BT and O3 lies on CT and that circle O1O2O3 passes through O, P and T. Thus, as is known, triangle O1O2O3 is similar to ABC and circles ABC, UVW, O1O2O3 are coaxal. Details of the proofs are left to the reader. 5
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References 1. F. E. Wood, Similar-Perspective triangles, Amer. Math. Monthly, 36:2 (1929) 67-73. 2. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, UK, 2007.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 6
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Article 59 7 Points on any Circle not through a Vertex Christopher J Bradley
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1. Introduction The general result (of which a particular case is illustrated above and analysed below in later sections) is as follows: Choose a point Q not on the sides or extensions of the sides of a triangle (in the illustration the centroid G is chosen), then points L, M, N are chosen on the sides BC, CA, AB so that Q is the Miquel point for circles AMN, BNL, CLM. Any appropriate choice of points 1
can be made. Any two points Y and Z are now chosen so that circle QYZ does not pass through a vertex (in the Figure Y and Z are such that BY = ¾ BG and CZ = ¾ CG). The three Miquel circles AMN, BNL, CLM now intersect circle QYZ again in points A', B', C' respectively. The beautiful result is that AA', BB', CC' are concurrent at a point P, which lies on circle QYZ. In what follows we establish the validity of the particular case, using areal co-ordinates. The general case is too difficult for DERIVE. It is hoped that a pure proof will emerge. 2. The circle GYZ The points Y and Z are chosen to have co-ordinates (1, 2, 1) and (1, 1, 2) respectively, so that BY = ¾BG and CZ = ¾CG. The circle through these points and G(1, 1, 1) has equation 3(b2 + c2)x2 + (2a2 – b2 + 2c2)y2 + (2a2 + 2b2 – c2)z2 – (8a2 – b2 – c2)yz + (2a2 – 7b2 + 2c2)zx + (2a2 + 2b2 – 7c2)xy = 0. (2.1) 3. The circle BLG and the points B' and N The choice of L is open to us and we choose it to lie at the midpoint of BC with co-ordinates (0, 1, 1). The equation of BLG now follows and is (a2 – 2b2 – 2c2)x2 – 3a2z2 + 3a2yz – 2(a2 – 2b2 + c2)zx + (a2 – 2b2 + 4c2)xy = 0. (3.1) The circles GYZ and BLG meet at G and again at the point B' with co-ordinates (x, y, z, where x = b2 + c2 – 2a2, y = 2c2 + 2a2 – b2, (3.2) z = 3c2. The circle BGL meets AB again at the point N with co-ordinates (a2 + 4c2 – 2b2, 2(b2 + c2) – a2, 0). 4. The circle ANG and the points A' and M Now that N is known we can derive the equation of circle ANG, which is (a2 – 2b2 + 4c2)y2 + (a2 + 4b2 – 2c2)z2 – 2(2a2 – b2 – c2)yz + (a2 – 2b2 – 2c2)(zx + xy) = 0. (4.1) Circle ANG and GYZ meet at G and again at the point A' with co-ordinates (x, y, z), where x = 4b2c2 – b4 – c4, y = b2(b2 + c2), (4.2) 2 2 2 z = c (b + c ). Circle ANG meets CA at M with co-ordinates (a2 – 2c2 + 4b2, 0, 2b2 + 2c2 – a2).
2
5. The circle CMG and the point C' Now that M is known we can derive the equation of circle CMG, which is (a2 – 2b2 – 2c2)x2 – 3a2y2 + 3a2yz + (a2 + 4b2 – 2c2)zx – 2(a2 + b2 – 2c2)xy = 0.
(5.1)
This circle, by Miquel circle theory, must pass through L and this has been checked. It also meets circle GYZ at the point C' with co-ordinates (b2 + c2 – 2a2, 3b2, 2a2 + 2b2 – c2). 6. AA', BB', CC' meet at a point P lying on circle GYZ It is now straightforward to show that AA', BB', CC' are concurrent at a point P whose coordinates are (b2 + c2 – 2a2, 3b2, 3c2) and that P lies on circle GYZ.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
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Article 60 Centroid-centred Similar Ellipses Christopher J Bradley
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1. Introduction The three ellipses drawn in the Figure are (i) the ellipse centre G passing through A, B, C known as the outer Steiner ellipse, (ii) the ellipse centre G touching the sides of ABC at its midpoints X, Y, Z and (iii) the intermediate ellipse centre G passing through points a quarter of the way along each side from each vertex, the points L, M, N, U, V, W. The co-ordinates in areals of some of the key points are listed and it is noted that AU, BM, CG amongst others are Cevian lines. It is also noted that if UG meets AB at H and MG meets AB at K, then HK is 1/5 AB. These ellipses are, of course, produced by an affinity from similar circles in an equilateral triangle. 1
2. The equations of the conics The outer Steiner ellipse is well known to have an equation yz + zx + xy = 0. (2.1) It is the circumscribing conic of minimum area. It meets the circumcircle at the Steiner point whose co-ordinates are (1/(b2 – c2), 1/(c2 – a2), 1/(a2 – b2)). The inner Steiner ellipse is well known to have the equation x2 + y2 + z2 – 2yz – 2zx – 2xy = 0. It is the inscribed conic of maximum area.
(2.2)
The inner and outer Steiner ellipses carry triangles that have an affinity with Poncelet’s porism. The conic through the points L(0, 3, 1), M(1, 0, 3), N(3, 1, 0), U(0, 1, 3), V(3, 0, 1), W(1, 3, 0) has equation 3x2 + 3y2 + 3z2 – 10yz – 10zx – 10xy = 0 (2.3) For obvious reasons we call this the intermediate Steiner conic. 3. A few properties of the intermediate Steiner conic The line UG has equation 3y = 2x + z, so it meets z = 0 at the point H(3, 2, 0) so that AH = 2/5 AB. The line UG meets the conic again at the point with co-ordinates (8, 5, – 1) and G(1, 1, 1) is the midpoint of this segment. It follows by a sequence of such arguments that G is the centre of the intermediate Steiner conic. The line MG meets AB at a point K such that AK = 3/5 AB. It follows that HK = 1/5 AB and that the midpoint of HK is the midpoint Z of AB. The lines AU, BM, CG meet at the point R (1, 1, 3). There are two other similar Cevian points with co-ordinates (3, 1, 1) and (1, 3, 1). All three conics are similar as they arise from the same affinity. Flat 4 Terrill Court 12-14, Apsley Road BRISTOL BS8 2SP
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Article 61 Concurrent lines in a Triangle with a Circle cutting the Sides Christopher J Bradley Z
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1. Introduction If ABC is a triangle and P a Cevian point such that AP, BP, CP meet BC, CA, AB respectively in L, M, N and then a circle is drawn through L, M, N to meet BC, CA, AB respectively at U, V, W, then it has been known for a long time, see Bradley [1], that AU, BV, CW are concurrent at 1
another Cevian point Q. When this construction is carried out a number of other sets of three lines are concurrent and sets of three points are collinear. In this article we exhibit some of these in the Figure above and in sections that follow we establish their validity, using areal coordinates with ABC as triangle of reference. The results we establish are as follows: (i) If VW meets BC at X, NL meets CA at Y and UM meets AB at Z then X, Y, Z are collinear; (ii) If NL meets UM at D, UM meets VW at E and VW meets NL at F, then AD, BE, CF are concurrent at a point R; (iii) MN, BE and LV are concurrent at a point G. Symmetry considerations now imply WU, MN, CF are concurrent at a point H and WU, AD, LV are concurrent at a point K. Permutation of letters L, M, N and U, V, W and A, B, C provide other collinearities and concurrences. 2.
The circle through L, M, N and the points U, V, W
Let P be the point (l, m, n) so that L(0, m, n), M(l, 0, n) and N(l, m, 0). The equation of the circle LMN is found to be mn(a2mn(l + m)(n + l) – b2nl(m + n)(l + m) – c2lm(n + l)(m + n))x2 + ... + ... – (m + n)(a2mn(l + m)(n + l) + l(n – m)(b2n(l + m) – c2m(n + l)))yz – ... – ... = 0. (2.1) This circle cuts BC, CA, AB at U, V, W whose co-ordinates are U: (0, a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)), a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l))); V: (b2nl(m + n)(l + m) – m(n + l)(c2l(m + n) – a2n(l + m)), 0, b2nl(m + n)(l + m) + m(n + l)(c2l(m + n) – a2n(l + m))); W: (a2mn(l + m)(n + l) – b2nl(l + m)(m + n) + c2lm(n + l)(m + n), – a2mn(l + m)(n + l) + b2nl(l + m)(m + n) + c2lm(n + l)(m + n), 0). 3.
AU, BV, CW are concurrent at Q
The equation of AU is (a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l)))y = (a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))z. (3.1) The equation of BV is (b2nl(m + n)(l + m) + m(n + l)(c2l(m + n) – a2n(l + m)))x = (b2nl(m + n)(l + m) – m(n + l)(c2l(m + n) – a2n(l + m)))z. (3.2) AU and BV meet at the point Q with co-ordinates (x, y, z), where x = 1/(l(m + n)(bn(l + m) + c2m(n + l)) – a2mn(l + m)(n + l)), y = 1/(a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l))), 2
(3.3)
z = 1/(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l))). It may now be checked that Q lies on the line CW. 4.
The lines NL, UM, WV and the collinear points X, Y, Z
The equation of the line NL is nly = mnx + lmz.
(4.1)
The equation of the line UM is n(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))x + l(a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l)))y + (a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))z. (4.2) The equation of the line WV is (a2(mn(l2 + lm + mn + nl) – b2nl(m2 + lm + mn + nl) – c2lm(n2 + lm + mn + nl)))x + (a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l)))y + (a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))z = 0. (4.3) NL meets CA at the point Y(– l, 0, n). UM meets AB at the point Z(–l(a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l))), n(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l))), 0). WV meets BC at the point X(0, a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)), – (a2mn(l + m)(n + l) – l(m + n)(b2n(l + m) – c2m(n + l)))). When the co-ordinates of X, Y, Z are put in the rows of a matrix, then the determinant of that matrix, as has been checked, vanishes, showing the points to be collinear. 5. The points D, E, F and then triangle ABC is in perspective with triangle DEF D is the intersection of NL and UM and has co-ordinates (x, y, z), where x = l(l(m + n)2(b2n(l + m) – c2m(n + l)) – a2mn(l + m)(n + l)(m – n)), y = 2mn (a2mn (l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l))), z = n(m + n)(a2mn(l + m)(n + l) + l(n – m)(b2n(l + m) – c2m(n + l))).
(5.1)
E is the intersection of UM and WV and has co-ordinates (x, y, z), where x = 2l(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))K, y = (a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))(a2mn(l3 + l2(m + 2n) + ln(2m + n) + mn2) – l(m + n)(b2n(l + m)(l – n) + c2(n + l)2)), (5.2) 3
z = (n + l)(l(m + n)(b2n(l + m) + c2m(l – n)) – a2mn(l2 + lm – nl – mn))K, K = l(m + n)(b2n(l + m) – c2m(n + l)) – a2mn(l + m)(n + l). F is the intersection of WV and NL and has co-ordinates (x, y, z), where x = l(m + n)(a2mn(l2 + lm + mn + nl) + l(n – m)(b2n(l + m) – c2m(n + l))), y = m(n + l)(l(m + n)(b2n(l + m) + c2m(l – n)) – a2mn(l2 + lm – nl – mn)), (5.3) 2 2 2 2 3 2 z = n(l(m + n)(b n(l + m) + c m(l – m)(l + n)) – a mn(l + l (2m + n) + lm(m + 2n) + m2n)). It may now be checked that the point R with co-ordinates (x, y, z) below lies on all three lines AD, BE, CF. x = 2l(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))H, y = 2m(n + l)(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l)))(l(m + n)(b2n(l + m) + c2m(l – n)) – a2mn(l2 + lm – nl – mn)), (5.4) 2 2 2 2 z = (n + l)(l(m + n)(b n(l + m) + c m(l – n)) – a mn(l + lm – nl – mn))H, H = (m + n)(a2mn(l + m)(n + l) + l(n – m)(b2n(l + m) – c2m(n + l))). 6. BE, MN, LV are concurrent at a point G The line MN has equation mnx = nly + lmz.
(6.1)
The line BE has equation ((n + l)(l(m + n)(b2n(l + m) + c2m(l – n)) – a2mn(l2 + lm – nl – mn)))x = (2l(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l))))z. (6.2) The line LV has equation (a2mn(l + m)(n + l) + b2nl(l + m)(m + n) – c2lm(n + l)(m + n))(ny – mz) – mx(a2mn(l + m)(n + l) – b2nl(l + m)(m + n) – c2lm(n + l)(m + n)) = 0. (6.3) It may now be checked that these three lines all pass through the point G with co-ordinates (x, y, z), where x = 2nl(a2mn(l + m)(n + l) + l(m + n)(b2n(l + m) – c2m(n + l))), y = m(a2mn(n + l)(l2 +lm + mn + nl) – l(m + n)(b2n(l + m)(l – n) + c2m(n + l)2)), (6.4) z = n(n + l)(l(m + n)(b2n(l + m) + c2m(l – n)) – a2mn(l2 + lm – nl – mn)). As mentioned in Section 1 symmetry considerations now imply WU, MN, CF are concurrent at a point H and WU, AD, LV are concurrent at a point K. Flat 4, 12-14, Apsley Road, BRISTOL BS8 2SP 4
Article 62 A Triangle with an Arbitrary Conic cutting its Sides Christopher J Bradley
H
Q P
Z S
R
F
N
A V E
W G
B
M L
U C
X
D K
Y
1
1. Introduction A conic is chosen that cuts the sides BC, CA, AB of a triangle ABC in points L, U; M, V; N, W (in that order anticlockwise) respectively. The chords LW, MU, NV form a triangle DEF as shown in the Figure. The following properties now hold: (i) AD, BE, CF are concurrent at a point P; (ii) If NV meets BC at X, LW meets CA at Y, MU meets AB at Z then X, Y, Z are collinear; (iii) NU, LV, AD are concurrent at a point Q; (iv) LV, MW, BE are concurrent at a point R; (v) MW, NU, CF are concurrent at a point S; (vi) If MW meets BC at G, NU meets CA at H, LV meets AB at K, then G, H, K are collinear. These results are illustrated in the Figure and are established as true in later Sections using areal co-ordinates with ABC as triangle of reference. 2. The choice of conic and the points L, M, N, U, V, W The main difficulty with the work in this article is the choice of equation of a general conic. If one of the usual forms is chosen than the points on the sides turn out to have square roots in them, which are virtually impossible to handle algebraically. Once this difficulty is solved so that the co-ordinates of the six points L, M, N, U, V, W have rational co-ordinates, then the results follow fairly easily. The equation of the conic chosen with a view to this end is vw(f – u)(g – v)(h – w)(gh – vw)x2 + ... + ... – (f2(gh + vw) – fu(gw + vh + 2vw) + 2u2vw)yz – ... – ... = 0. (2.1) The co-ordinates of the points on the sides may now be calculated and are: L : (0, v(f – u), hf – wu); M : (fg – uv, 0, w(g – v)); N : (u(h – w), gh – vw, 0); U : (0, fg – uv, w(f – u)), V : (u(g – v), 0, gh – vw); W : (hf – wu, v(h – w), 0).
(2.2)
Carnot’s theorem (see Bradley [1]), (BL/LC)(BU/UC)(CM/MA)(CV/VA)(AN/NB)(AW/WB) = 1
(2.3)
follows immediately 3. Points D, E, F and the perspective of ABC and DEF The co-ordinates of the points D, E, F are D(f, v, w), E(u, g, w), F(u, v, h) so that AD, BE, CF are concurrent at the point P(u, v, w). 4. The points X, Y, Z and their collinearity 2
The equation of the line DE is w(v – g)x + w(u – f)y + (fg – uv)z = 0.
(4.1)
The equations of the lines EF and FD now follow by cyclic change of x, y, z and f, g, h and u, v, w. EF meets BC at X(0, g – v, w – h); Y is (u – f, 0, h – w), and Z is (f – u, v – g, 0). It follows that X, Y, Z are collinear and the equation of XYZ is x/(u – f) + y/(v – g) + z/(w – h) = 0. (4.2) 5. LV, NU and AD are concurrent The equation of LV is v(f – u)(gh – vw)x + u(g – v)(fh – wu)y + uv(u – f)(g – v)z = 0.
(5.1)
The equation of NU is w(f – u)(gh – vw)x + wu(h – w)(f – u)y + u(h – w)(uv – fg)z = 0.
(5.2)
These meet at the point Q(fu(g – v)(w – h), v(f – u)(gh – vw), w(f – u)(gh – vw)) and this obviously lies on AD with equation wy = vz. Similarly results (iv) and (v) hold. 6. The collinear points G, H, K LV meets AB at K(u(v – g)(fh – wu), v(f – u)(gh – vw), 0). NU meets CA at H(u(w – h)(fg – uv), 0, w(f – u)(gh – vw)). The equation of KH is vw(f – u)(gh – vw)x + wu(g – v)(hf – wu)y + uv(h – w)(fg – uv)z = 0. (6.1) The symmetry of this equation implies this line also passes through the point G = BC^MW. Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, UK, (2007). Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article 63 A Cascade of Conics Christopher J Bradley
A
F N
V E
W
B
M
P
C
U
L
D Figure
1. Introduction Triangle ABC is given and a conic S is drawn through the vertices. A point P is chosen not on the sides or their extensions and AP, BP, CP are drawn to meet S again at D, E, F. ABC and DEF are therefore in perspective. DE is drawn to meet BC at U and CA at M. EF is drawn to meet CA at V and AB at N. FD is drawn to meet AB at W and BC at L. It now turns out that a conic Σ may be drawn through the six points L, U, M, V, N, W. Furthermore triangles LMN and VWU are in perspective with vertex P. It then follows by the same reasoning as before that the intersections of non-corresponding sides of these two triangles lie on a conic. The process may 1
be repeatedly indefinitely, producing the cascade of conics in the title. The result is proved in the following sections using homogeneous projective co-ordinates with ABC as triangle of reference and P as unit point. 2.
The conic S through A, B, C, D, E, F
Suppose the conic S through ABC has equation fyz + gzx + hxy = 0.
(2.1)
With P as unit point the line AP has equation y = z. It meets S at the point D with co-ordinates (– f, g + h, g + h). Similarly E and F have co-ordinates E(h + f, – g, h + f), F(f + g, f + g, – h ). 3. The points U, M, V, N, W, L The equation of the line DE is (g + h)x + (h + f)y – hz = 0.
(2.2)
DE meets BC at U(0, h, h + f) and DE meets CA at M(h, 0, g + h). Similarly V(f + g, 0, f), N(h + f, , f, 0), W(g, g + h, 0) and L(0, f + g, g). 4. The conic Σ through U, M, V, N, W, L The conic Σ may now be calculated and has equation f(g + h)x2 + g(h + f)y2 + h(f + g)z2 – (f2 + fg + hf + 2gh)yz – (g2 + fg + gh + 2hf)zx – (h2 + hf + gh + 2fg)
(2.3)
It may now be checked that WM, UN, VL all pass through P and so we once again have six points on a conic whose vetices form two triangles in perspective and so the whole process may be repeated and the cascade of conics is created.
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP
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Article 64 Porism created by the Circumcircle and Triangles in Perspective Christopher J Bradley
af
M
W
A ea
F
E
fb P
L
ce
B bd
C D
dc
N V
Figure
1. Introduction
1
U
Given a triangle ABC and its circumcircle S any point P is chosen that is not on its sides or the extension of the sides and lines AP, BP, CP are drawn to intersect S again at D, E, F respectively. The tangents at A, B, C meet EF, FD, DE at U, V, W respectively. The tangents at D, E, F meet BC, CA, AB at L, M, N respectively. The tangents at A and F meet at af and points fb, bd, dc, ce, ea are similarly defined. The following results now hold: (i) U, V, W are collinear; (ii) L, M, N are collinear; (iii) points af, fb, bd, dc, ce, ea lie on a conic Σ. This fact and the construction leading up to it mean that the hexagon af fb bd dc ce ea has it vertices on Σ and its sides touch S. A hexagonal porism is thereby constructed. Furthermore (iv) af dc, fb ce, bd ea all pass through P so the details of the construction may be repeated ad infinitum, first with Σ as inconic and a new conic as circumconic. In the following Sections these results are established using areal co-ordinates with ABC as triangle of reference. 2. The points D, E, F Let P have co-ordinates P(l, m, n). The line AP has equation ny = mz and this meets the circumcircle S with equation a2 yz + b2zx + c2xy = 0 at the point D with co-ordinates (x, y, z), where x = – a2mn, y = m(b2n + c2m), z = n(b2n + c2m). (2.1) The co-ordinates of E, F may now be obtained by cyclic change of x, y, z and a, b, c and l, m, n. 3. The points U, V, W and their collinearity Given the co-ordinates of E, F from Section 2 we may obtain the equation of EF and this is l(y(a2n + c2l) + z(a2m + b2l)) – a2mnx = 0. (3.1) 2 2 The equation of the tangent to S at A is b z + c y = 0. These two lines meet at the point U with co-ordinates U(l(c2m – b2n), – b2mn, c2mn) The co-ordinates of the points V, W may be obtained from those of U, by cyclic change of x, y, z and a, b, c and l, m, n. The 3 x 3 determinant having as its rows the co-ordinates of U, V, W vanishes proving that U, V, W are collinear. By symmetry it follows (by reversing the roles of A, B, C and D, E, F) that L, M, N are also collinear. 4. The points af, fb, bd, dc, ce, ea and the conic Σ The equation of the tangents at A, B, C are b2z + c2y = 0, c2x + a2z = 0, a2y + b2x = 0 respectively. The tangent at D has equation (b2n + c2m)2x + a2b2n2y + c2a2m2z = 0. (4.1)
2
The tangents at E, F have equations that can be written down from Equation (4.1) by cyclic change of x, y, z and a, b, c and l, m, n. It is now possible to obtain the co-ordinates of af, fb, bd, dc, ce, ea, which are: af : (a2m + 2b2l, b2m, – c2m); fb : (a2l, 2a2m + b2l, – c2l); bd: (– a2n, b2n + 2c2m, c2n); dc: (– a2m, b2m, 2b2n + c2m); ce : (a2l, – b2l, 2a2n + c2l); ea : (a2n + 2c2l, – b2n, c2n). It s now an extensive calculation to show these six points lie on the conic Σ with equation a4b2c2m2n2(b2n + c2m)x2 +... + ...+ a2(a6m2n2(b2n + c2m) + 2a4lmn(b4n2 + 2b2c2mn + c4m2) + a2l2(b6n3 + 4b4c2mn2 + 4b2c4m2n + c6m3) + b2c2l3(b4n2 + c4m2))yz + ... + ... = 0. (4.1) The hexagon with the six points above has its vertices lying on the conic Σ and its sides touch the circumcircle S. Thus a porism of hexagons is created. Furthermore, it can be checked that fb P ce are collinear as are af P dc and bd P ea. Thus triangles af bd ce and fb dc ea have vertices on Σ and are in perspective. It follows that the construction may be repeated indefinitely. See Article 28 for further information about porisms created by a pair of triangles in perspective.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 65 When I replaces K and Ge replaces H and Mi replaces O Christopher J Bradley
A
W T'
F T
X'
E' S'
Z W L
F'
Mi
M
G
Ge
N' I N W'
V
Y S E Z'
Y' R B
X
D
D'
C
R' U
1. Introduction A study is made of what happens when the incentre I with areal co-ordinates (a, b, c) replaces the symmedian point K with areal co-ordinates (a2, b2, c2) in the construction of the Triplicate Ratio 1
Circle and the 7-point circle, and indeed in the circumcircle and the nine-point circle. Other points whose areal co-ordinates depend on a2, b2, c2 only get replaced by those with identical functions of a, b, c. So, for example Gergonne’s point Ge (1/(b + c – a), 1/(c + a – b), 1/(a + b – c)) replaces H (1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)) and the Mittenpunkt Mi replaces the circumcentre O. Details are shown in the Figure and are explained more fully in the paragraphs that follow. An affine transformation is involved, so that midpoints remain as midpoints (and hence the centroid G and the midpoints of sides X, Y, Z remain unchanged). There is nothing really new in this article, but the various comparisons and constructions may well be a novelty to the reader. It is a classic example of how to obtain seemingly new results from those existing in the Euclidean plane. 2. The circumconic Take lines AGe, BGe, CGe to met BC, CA, AB respectively in R, S, T and on these lines choose points R', S', T' so that GeR = RR', GeS = SS', GeT = TT'. The circumconic ABCR'S'T' may now be drawn and it has equation ayz + bzx + cxy = 0. (2.1) Its centre is the Mittenpunkt Mi with co-ordinates (a(b + c – a), b( c + a – b), c(a + b – c)). Note that G remains in position with co-ordinates (1, 1, 1) and since Mi replaces O and Ge replaces H it follows that Mi, G, Ge are collinear and MiG = (1/3)MiGe. 3. The nine-point conic In keeping with the classical construction the nine points on this conic are the feet R. S, T of the Cevians through Ge, the midpoints X, Y, Z of the sides BC, CA, AB respectively and the midpoints X', Y', Z' of the line segments AGe, BGe, CGe respectively. Its centre is the point N', the midpoint of MiGe. 4. The triplicate ratio conic In the standard construction of the triplicate ratio circle one starts with the symmedian point K and draws the lines through K parallel to the sides. Here we use the incentre I, rather than K and then the lines through I parallel to the sides define the points D, D', E, E', F, F' with co-ordinates: D(0, a + b, c), D'(0, b, c + a), E(a, 0, b + c), E'(a + b, 0, c), F(c + a, b, 0), F'(a, b + c, 0). The conic through these six points is the triplicate ratio conic with equation (b + c)x2/a + (c + a)y2/b + (a + b)z2/c – yz(2 + a(a + b + c)/bc) – zx(2 + b(a + b + c)/ca) – xy(2 + c(a + b + c)/ab) = 0. (4.1)
2
If the lines F'D, D'E, E'F are drawn these lines define a triangle UVW such that U, V, W lie on the circumcircle of triangle ABC. 5. The 7-point conic Lines through A, B, C parallel to the sides FD, DE, EF respectively concur at the Brocard like point W (1/b, 1/c, 1/a) and lines through A, B, C parallel to the sides D'E', E'F', F'D' concur at the Brocard like point W'(1/c, 1/a, 1/b). Let BW^CW' = L, CW^AW' = M, AW^BW' = N. L lies also on both MiX and F'E and has co-ordinates (a, c, b). M lies also on both MiY and D'F and has coordinates (c, b, a). N lies also on both MiZ and E'D and has co-ordinates (b, a, c). The seven points Mi, W, W', L, M, N, I lie on the 7-point conic with the equation bcx2 + cay2 + abz2 – a2yz – b2zx – c2xy = 0 (5.1) The triplicate ratio conic and the 7-point conic have identical centres, the midpoint of IMi. 6. The Brocard like conics It is well known that an additional property of the Brocard points Ω, Ω' is that there are equivalent defining properties involving there circles through Ω and three circles through Ω'. We show in the figure only one of these, which is an ellipse that touches AB at B and passes through C and W' and has its centre on the line LMiX. There are six of these ellipses, three passing through W and three passing through W'.
Flat 4, Terrill Court, 12-14, Apsley Road, Bristol BS8 2SP
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Article 66 On the mean of two Cevian points Christopher J Bradley
A W F E
V
K S P Q
N J B
D
L
R
M
C
U
Figure
1. Introduction Given two Cevian points there are many ways of defining their mean, the most obvious being the midpoint of the line joining them. But the question is whether some particular choice has any geometrical significance. In this short article we provide a construction that leads to a point with some particular significance, but it is not suggested it is the only reasonable choice. 1
We use the fact that the six feet of the pair of Cevian points always lies on a conic. If one defines points L, D , M, E, F, N (in that order) on the sides BC, CA, AB respectively (so that L, M , N are the feet of the Cevian point J and D, E, F are the feet of the Cevian point K) then the lines NL, DM, EF form a triangle UVW (see the Figure above) and in this article we show that AU, BV, CW are concurrent at a point P. Other possibilities might be chosen, as there are numerous other concurrencies in the figure, which we also investigate. 2. The points U, V, W and the Cevian mean If J has co-ordinates (u, v, w) then L, M, N have co-ordinates L(0, v, w), M(u, 0, w) and N(u, v, 0). Similarly if K has co-ordinates (l, m, n) then D, E, F have co-ordinates D(0, m, n), E(l, 0, n) and F(l, m, 0). The equation of NL is therefore wuy – uvz – vwx = 0.
(2.1)
The equation of DM is umz – mwx – uny = 0.
(2.2)
And the equation of EF is nly + lmz – mnx = 0.
(2.3)
The point U = NL^DM and therefore has co-ordinates U(u(mw – nv), 2mvw, w(mw + nv)). The point V = EF^DM and therefore has co-ordinates V(2nlu, m(nu – lw), n(lw + nu)). The point W = NL^EF and therefore has co-ordinates W(lu(mw + nv), mv(lw + nu), nw(mu – lv)). It is now straightforward to show that AU, BV, CW are concurrent at a point P with co-ordinates P(2lu(mw + nv), 2mv(lw + nu), (lw + nu)(mw + nv)). The fact that the co-ordinates of P are not symmetrical leads one to consider why, and the reason is obvious. One can perform the same analysis with chords LM, EN, FD creating another triangle U'V'W' and then AU', BV', CW' are concurrent at the point P' with co-ordinates that may be written down from those of P with cyclic change of x, y, z and l, m, n and u, v, w. And then one can perform the same analysis with the chords MN, FL, DE creating a third triangle U''V''W'' and another point P'' with co-ordinates that again may be written down by a further set of cyclic changes. Possibly the centroid of the triangle PP'P'' is a candidate as the mean of the Cevian points J and K. Its co-ordinates are (x, y, z), where, x = 5lu(mw + nv) + mnu2 + vwl2 etc. 3. Points Q, R, S
2
It may be shown that FD, LE, AU are concurrent at the point Q with co-ordinates (l(nv – mw), 2mnv, n(mw + nv)) and that DE, MF, BV are concurrent at a point R with co-ordinates that may be written down by cyclic change of x, y, z and u, v, w and l, m, n. Similarly S = EF^ ND^CW. Many other sets of concurrent lines exist, but there is little point in recording them.
Flat 4 Terrill Court 12-14 Apsley Road, BRISTOL BS8 2SP
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Article 67 A Special Tucker Circle Christopher J Bradley
A 23
W
V 32
F 13
L Q
K P
R E
H
M N
12 B
31`
D
21
C
U
1. Introduction Suppose we are given a triangle ABC, orthocentre H and feet of the altitudes D, E, F on BC, CA, AB respectively. From D perpendiculars are drawn to CA, AB meeting them at points 12 and 13 respectively. Points 21 and 23 lie on BC, AB respectively where the perpendiculars from E to those sides meet them. Similarly points 31, 32 lie on BC, CA respectively where the perpendiculars from F to those sides meet them. The lines 13 31, 12 21, 23 32 are now drawn creating a triangle UVW, where U = 13 31^12 21, V = 21 12^23 32 and W = 13 31^32 23. The lines 12 13, 23 21, 31 32 are also drawn. See the Figure above. 1
The following results now hold: (i) A Tucker circle passes through the six points 23 13 31 21 12 32 with 23 32 parallel to BC, 31 13 parallel to CA and 12 21 parallel to AB; (ii) AU, BV, CW are concurrent at the symmedian point K; (iii) Lines 21 23, 31 32, AKU are concurrent at a point L; (iv) Lines 13 12, 31 32, BKV are concurrent a point M; (v) Lines 21 23, 12, 13, CKW are concurrent at a point N. CABRI also indicates that circles ABC, the Tucker circle and circle UVW have their centres Q, P, R lying on a line with P the midpoint of QR. The collinearity of these points is proved in Bradley [1]. In the next sections results (i) to (v) are proved using areal co-ordinates with ABC as triangle of reference. 2. The co-ordinates of the six points We take the co-ordinates of H to be H(u, v, w). When this is done it is well known that the coordinates of K, the symmedian point, are K(u(v + w), v(w + u)) and Q, the circumcentre of ABC are Q(v + w, w + u, u + v). The feet of the altitudes now have co-ordinates D(0, v, w), E(u, 0, w) and F(u, v, 0). The line parallel to BH through D has equation w(u + v + w)x + wuy – uvz = 0. (2.1) The line parallel to CH through D is v(u + v + w)x – wuy + uvz = 0.
(2.2)
The point 12 is where the line with Equation (2.1) meets CA, y = 0, and so has co-ordinates 12(uv, 0, w(u + v + w)). The point 13 is where the line with Equation (2.2) meets AB, z = 0, and so has co-ordinates 13(wu, v(u + v + w), 0). The other four points may now be witten down by cyclic change of x, y, z and u, v, w. Thus 23 has co-ordinates 23(u(u + v + w), vw, 0), 21 has coordinates 21(0, uv, w(u + v + w)), 31 has co-ordinates (0, v(u + v + w), wu) and 32 has coordinates 32(u(u + v + w), 0, vw). 3. The equation of the special Tucker circle First note that the co-ordinates of 23 and 32 imply that the line 23 32 divides AB and AC in the same ratio and hence the line 23 32 is parallel to BC. Similarly 31 13 is parallel to CA and 12 21 is parallel to AB. The equation of the circle through the six points is v2w2(u + v + w)x2 + ... + ... – vw(2u2 + 2u(v + w) + (v + w))yz – ... – ... = 0.
2
(3.1)
4. The points U, V, W and the concurrence of AU, BV, CW The equation of the line 31 13 is v(u + v + w)(z + x) = wuy.
(4.1)
The equation of the line 12 21 is w(u + v + w)(x + y) = uvz.
(4.2)
The equation of the line 23 32 is u(u + v + w)(y + z) = vwx. The point U = 12 21^31 13 and therefore has co-ordinates U(– vw(2u + v + w), v(w + u)(u + v + w), w(u + v)(u + v + w)).
(4.3)
(4.4)
The co-ordinates of the points V, W may now be written down from (4.4) by cyclic change of x, y, z and u, v, w and are V(u(v + w)(u + v + w), – wu(2v + w + u), w(u + v)(u + v + w)), (4.5) and W(u(v + w)(u + v + w), v(w + u)(u + v + w), – uv(2w + u + v)).
(4.6)
It is obvious from the co-ordinates of U, V, W that AU, BV, CW all pass through the symmedian point K(u(v + w), v(w + u), w(u + v)). This is also the symmedian point of triangle UVW. 5. The circumcircle of triangle UVW The equation of the circle UVW may now be obtained and is vw(u + v + w)(u2 + u(v + w) – 2vw)x2 + ... + ... – u(u2(v2 – 4vw + w2) + u(v + w)(v2 – 6vw + w2) – 3vw(v + w)2)yz – ... – ... = 0.
(5.1)
6. The concurrence of AU, 23 21, 32 31 The equation of AU is w(u + v)y = v(w + u)z. The equation of the line 23 21 is vw2x = u(w(u + v + w)y – uvz). The equation of the line 32 31 is v2wx + u(wuy – v(u + v + w)z) = 0. 3
(6.1)
(6.2)
(6.3)
These three lines have a point L in common with co-ordinates L(u(2u + v + w), v(w + u), w(u + v))
(6.4)
Results (iv) and (v) follow immediately by cyclic change of letters L, M, N and x, y, z and u, v, w.
References 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, (2007).
Flat 4 Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 68 Six Collinear Points in a Special Cyclic Quadrilateral Christopher J Bradley
W
A 1 Z
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B
P
J
S E da
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ab T
To X
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R
bc X
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1. Introduction This article describes the special properties of a cyclic quadrilateral ABCD, centre O, when the diagonals AC and BD cut at right angles at a point E. It is found that four additional points lie on OE. Most, if not all, of the results described here are well known. It is indeed well known that when AC and BD are at right angles that the tangents at A, B, C, D create a quadrilateral WXYZ that is cyclic. Circle WXYZ has centre the point V and V lies on 1
OE. Circles AOB, BOC, COD, DOA have centres denoted by ab, bc, cd, da and these centres turn out to be concyclic with centre the point K. The point K also lies on OE and is the midpoint of OV. If P, Q, R, S are the midpoints of AB, BC, CD, DA respectively then PR and QS intersect at a point T and T also lies on OE. Also T is the centre of the so-called 8-point circle passing through P, Q, R, S and 1, 2, 3, 4. Here 1, 2, 3, 4 lie on AB, BC, CD, DA respectively and are the points where perpendiculars from the midpoints of opposite sides meet these lines. So, for example 1 is the point on AB such that R1 is perpendicular to AB. The four lines R1, S2, P3, Q4 also pass through E. And T is the midpoint of OE. Finally the lines 13, 24 intersect at the point J and J lies on OE. See the figure above that illustrates all these results. In subsequent sections we prove these results using Cartesian coordinates. In Bradley [1] the results involving the 8-point circle are given in greater detail and some proofs there are pure, rather than analytic. 2. The points A, B, C, D and E We take circle ABCD to be the unit circle, so that A, B, C, D have co-ordinates of the form ((1 – p2/(1 + p2), 2p/(1 + p2)). Since the diagonals of ABCD intersect at right-angles, suitable axes are along the bisectors of angle AOC and BOD respectively and then since A and C exchange under refection in the x-axis and B and D exchange under reflection in the y-axis we may take A, B, C, D to have parameters p = – t, s, t, 1/s respectively. The advantage of this is that we can now make do with two parameters s and t rather than four parameters. The equation of AC is therefore (1 + t2)x = 1 – t2,
(2.1)
and the equation of BD is (1 + s2)y = 2s.
(2.2)
The co-ordinates of E are obviously E((1 – t2)/(1 + t2), 2s/(1 + s2)). We can now find the equations of the sides of ABCD and these are: AB: (1 + st)x + (s – t)y = 1 – st, BC: (1 – st)x + (s + t)y = 1 + st, CD: (s – t)x + (1 + st)y = s + t, DA: (s + t)x + (1 – st)y = s – t. 3. Circles AOB, BOC, COD, DOA, their centres ab, bc, cd, da and the point K 2
(2.3) (2.4) (2.5) (2.6)
The centre ab of circle AOB has co-ordinates ((1 + st)/(2(1 – st)), (s – t)/((2(1 – st))). The centre bc of circle BOC has co-ordinates ((1 – st)/(2(1 + st)), (s + t)/(2(1 + st))).
(3.1)
(3.2)
The centre cd of circle COD has co-ordinates ((s – t)/(2(s + t)), (1 + st)/(2(s + t))). The centre da of circle DOA has co-ordinates ((s + t)/(2(s – t)), (1 – st)/(2(s – t))).
(3.3)
(3.4)
The points ab, bc, cd, da are concyclic and lie on the circle with equation 8(s2 – t2)(s2t2 – 1)(x2 + y2) – 2(1 + s2)2(t4 – 1)x + 4s(1 + s2)(1 + t2)2y – (1 + s2)2(1 + t2)2 = 0. (3.5) It may be seen that the centre K of this circle has co-ordinates (u, v), where u = k(1 – t2)/(1 + t2), v = 2ks/(1 + s2) and k = (1 + s2)2(1 + t2)2/(8(s2 – t2)(1 – s2t2)).
(3.6)
It is evident that K lies on OE with equation (2s/(1 + s2))x = ((1 – t2)/(1 + t2))y. 4. Points P, Q, R, S and circle PQRS centre T P is the midpoint of AB and has co-ordinates ((1 – s2t2)/((1 + s2)(1 + t2)), (s – t)(1 – st)/((1 + s2)(1 + t2))).
(4.1)
Q is the midpoint of BC and has co-ordinates ((1 – s2t2)/((1 + s2)(1 + t2)), (s + t)(1 + st)/((1 + s2)(1 + t2))).
(4.2)
R is the midpoint of CD and has co-ordinates ((s2 – t2)/((1 + s2)(1 + t2)), (s + t)(1 + st)/((1 + s2)(1 + t2))).
(4.3)
S is the midpoint of DA and has co-ordinates ((s2 – t2)/((1 + s2)(1 + t2)), (s – t)(1 – st)/((1 + s2)(1 + t2))).
(4.4)
The equation of the circle PQRS, the so-called 8-point circle, is (1 + s2)2(1 + t2)2(x2 + y2) – (1 – t4)(1 + s2)2x – 2s(1 + s2)(1 + t2)2y + 2(1 – s2t2)(s2 – t2) = 0. (4.5) Its centre T lies on OE and is such that OT = TE. It may also be checked that T is the midpoint of both PR and QS, so that these lines are diameters of circle PQRS. T is also the intersection of 3
ab cd and bc da.The other 4 points on this circle are defined in the next section. 5. The points 1, 2, 3, 4 and the point J The line PE has equation (1 + s2)(1 + t2)(1 + st)x – (s – t)(1 + s2)(1 + t2)y – (1 – st)(s2t2 – s2 – t2 + 4st + 1) = 0. (5.1) This is perpendicular to CD and meets CD at the point 3 with co-ordinates (x, y), where x = – (1/((1 + s2)2(1 + t2)2)(s4(t4 – 2t2 – 1) + 4s3t3 – 4st + t4 + 2t2 – 1), y = (2/((1 + s2)2(1 + t2)2)(s4t3 + s3(3t2 + 1) – s2t(1 + t2) + st2(3 + t2) + t). (5.2) It may now be checked that 3 lies on circle PQRS. The line RE has equation (s – t)(1 + s2)(1 + t2)x – (1 + st)(1 + s2)(1 + t2)y + (s + t)(s2t2 – s2 – t2 + 4st + 1) = 0.
(5.3)
This is perpendicular to AB and meets AB at the point 1 with co-ordinates (x, y), where x = – (1/((1 + s2)2(1 + t2)2)(s4(t4 + 2t2 – 1) + 4st(s2 – t2) + t4 – 2t2 – 1), y = – (2/((1 + s2)2(1 + t2)2(s4t – s3t2(3 + t2) – s2t(1 + t2) – s(1 + 3t2) + t3).
(5.4)
It may now be checked that 1 lies on circle PQRS. The line 13 has equation (s2 + 1)(s4 – 1)(t2 + 1)3x – 2t(s2 + 1)3(t2 + 1)2y + (st – t + s + 1)(st + t – s + 1)((s4 + 1)(t4+ 4t2 + 1) + 4s2t2) = 0. (5.5) This meets the line OE at the point J with co-ordinates (x, y), where x = m(1 – t2)/(1 + t2), y = 2sm/(1 + s2), m = ((s4 + 1)((t4 + 4t2 + 1) + 4s2t2))/ ((1 + s2)2(1 + t2)2) .
(5.6)
The line QE has equation (1 + s2)(1 + t2)(st – 1)x + (1 + s2)(1 + t2)(s + t)y + (1 + st)((1 – s2)(1 – t2) – 4st).
(5.7)
This line is perpendicular to DA and meets DA at the point 4 with co-ordinates (x, y), where x = – (1/(1 + s2)2(1 + t2)2)(s4(t4 – 2t2 – 1) – 4st(s2t2 – 1) + t4 + 2t2 – 1), y = – (2/(1 + s2)2(1 + t2)2)(s4t3 – s3(1 + 3t2) – s2t(1 + t2) – st2(3 + t2) + t). (5.8) It may now be checked that 4 lies on circle PQRS. The line SE has equation 4
(s + t)(1 + s2)(1 + t2)x + (st – 1)(1 + s2)(1 + t2)y + (s – t)((s2 – 1)(t2 – 1) – 4st).
(5.9)
This line is perpendicular to BC and meets BC at the point 2 with co-ordinates (x, y), where x = – (1/(1 + s2)2(1 + t2)2)(s4(t4 + 2t2 – 1) – 4st(s2 – t2) + t4 – 2t2 – 1), y = – (2/(1 + s2)2(1 + t2)2)(s4t + s3t2(3 + t2) – s2t(1 + t2) + s(1 + 3t2) + t3). (5.10) It may now be checked that 2 lies on circle PQRS and that the line 14 passes through J. 6. The cyclic quadrilateral formed by the tangents at A, B, C, D The equations of the tangents at A, B, C, SD are as follows. At A: (1 – t2)x – 2ty = 1 + t2, At B: (1 – s2)x + 2sy = 1 + s2, At C: (1 – t2)x + 2ty = 1 + t2, At D: (s2 – 1)x + 2sy = 1 + s2.
(6.1) (6.2) (6.3) (6.4)
The tangents at A and B meet at W with co-ordinates ((1 + st)/(1 – st), (s – t)/(1 – st)). The tangents at B and C meet at X with co-ordinates ((1 – st)/(1 + st), (s + t)/(1 + st)). The tangents at C and D meet at Y with co-ordinates ((s – t)/(s + t), (1 + st)/(s + t)). The tangents at D and A meet at Z with co-ordinates ((s + t)/(s – t), (1 – st)/(s – t)). It may now be shown that points X, Y, Z, W lie on a circle whose centre is the point V with coordinates (n(1 – t2)/(1 + t2), 2ns/(1 + s2), where n = (1 + s2)2(1 + t2)2/(4(s2 – t2)((1 – s2t2)). This shows that V lies on OE and is such that OK = KV. It may also be shown that XZ meets YW at the point T. Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath (2007).
Flat 4 Terrill Court 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 69 Two Cyclic Quadrilaterals and Two Coaxal Systems Christopher J Bradley
P3 S0 X R1
T0
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A' E
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R3 C C'
Q0 Q1 Figure
1. Introduction Let ABCD and A'B'C'D' be two cyclic quadrilaterals inscribed in the same circle T, centre O, but having the following properties: (i) their diagonals are at right angles, the former at E and the
1
latter at F; (ii) AC, BD are respectively parallel to A'C', B'D'. These conditions are essential for the consequential properties we now list. See the figure above. (i) AA', BB', CC', DD' form a quadrilateral T0= P0Q0R0S0 (with AA' meeting BB' at P0 etc.) which turns out to be cyclic, centre O0; (ii) AB', BC', CD', DA' form a quadrilateral T1= P1Q1R1S1(with AB' meeting BC' at P1etc.) which again turns out to be cyclic, centre O1; (iii)AC', BD', CA', DB' form a quadrilateral T2= P2Q2R2S2(with AC' meeting BD' at P2etc.) which again turns out to be cyclic, centre O2; (iv) AD', BA', CB', DC' form a quadrilateral T3= P3Q3R3S3(with AD' meeting BA' at P2etc.) which again turns out to be cyclic, centre O3; (v) O, O1, O3 are collinear; (vi) O, O0, O2, G are collinear, where G is the midpoint of EF; (vii) T, T1, T3 are coaxal, with common chord EF; (viii) T, T0, T2 are coaxal, with limiting point G. 2. Choice of axes and co-ordinates of points We use Cartesian co-ordinates and because they meet at right-angles the best choice of axes are the internal bisectors of angles AOC and BOD, and these are also the internal bisectors of angles A'OC' and B'OD' since AC, BD are parallel to A'C'. B'D' respectively. An arbitrary point on circle T is taken to have co-ordinates ((1 – p2)/(1+ p2), 2p/(1 + p2)), where p is an appropriate parameter. The parameters are chosen as follows: A –t, B s, C t, D 1/s, A'– m, B' n, C' m, D' 1/n. 3. The circle T0 The lines involved have equations AA': (1 – mt)x – (m + t)y = 1 + mt, BB': (1 – sn)x + (s + n)y = 1 + sn, CC': (1 – mt)x + (t + m)y = 1 + mt, DD': –(1 – sn)x + (n + s)y = 1 + sn.
(3.1) (3.2) (3.3) (3.4)
Points involved have co-ordinates (x, y) P0 = AA'^BB': x = – (mns + mnt +mst + nst + m + n + s + t)/(mns + mnt + mst + nst – m – n – s – t), y = 2(mt – ns)/(mns + mnt + mst + nst – m – n – s – t);
(3.5)
Q0 = BB'^CC': x = – (mns – mnt – mst + nst + m – n – s + t)/(mns – mnt – mst + nst – m + n + s – t), y = 2(ns – mt)/(mns – mnt – mst + nst – m + n + s – t);
(3.6)
2
R0 = CC'^DD': x = (mns – mnt – mst + nst +m – n – s + t)/(mns + mnt + mst + nst – m – n – s – t), y = 2(mnst – 1) )/(mns + mnt + mst + nst – m – n – s – t);
(3.7)
S0 = DD'^AA': x = (mns + mnt +mst + nst + m + n + s + t)/(mns – mnt – mst + nst – m + n + s – t), y = 2(1 – mnst)/(mns – mnt – mst + nst – m + n + s – t).
(3.8)
It may now be checked that P0, Q0, R0, S0 lie on a circle T0 with centre O0 co-ordinates (u, v) where, u = (1 + n2)(1 + s2)(m2t2 – 1)/{(mns + mnt + mst + nst – m – n – s – t)(mns – mnt – mst + nst – m + n + s – t)}, v = (1 + m2)(1 + t2)(n + s)(1 + ns)/{(mns + mnt + mst + nst – m – n – s – t)(mns – mnt – mst + nst – m + n + s – t)}. (3.9) 4. The circle T1 The lines involved have equations AB': (1 + nt)x + (n – t)y = 1 – nt, BC': (1 – ms)x + (m + s)y = 1 + ms, CD': (n – t)x + (1 + nt)y = n + t, DA': (m + s)x + (1 – ms)y = s – m.
(4.1) (4.2) (4.3) (4.4)
Points involved have co-ordinates (x, y) P1 = AB'^BC': x = – (mns + mnt – mst + nst – m + n – s – t)/(mns + mnt – mst + nst + m – n + s + t), y = 2(ms + nt)/(mns + mnt – mst + nst + m – n + s + t);
(4.5)
Q1 = BC'^ CD': x = – (mnst – mn + ms – mt + nt – ns – st + 1)/(mnst + mn + ms – mt + ns – nt – st – 1), y = 2(mns – t)/(mnst + mn + ms – mt + ns – nt – st – 1);
(4.6)
R1 = CD'^ DA': x = (mns – mnt + mst + nst – m – n + s – t)/(mns + mnt – mst + nst + m – n + s + t), y = 2(mn + st)/(mns + mnt – mst + nst + m – n + s + t);
(4.7)
S1 = DA'^AB': x = – (mnst + mn – ms – mt – ns – nt + st)/(mnst + mn + ms – mt + ns – nt – st – 1), y = 2(m – nst)/(mnst + mn + ms – mt + ns – nt – st – 1).
(4.8)
3
It may now be checked that P1, Q1, R1, S1 lie on a circle T1 with centre O1 co-ordinates (u, v) where, u = – (1 + m2)(1 + t2)(1 – sn)(s – n)/{(mns + mnt – mst + nst + m – n + s + t)(mnst + mn + ms – mt + ns – nt – st – 1)}, v = (m2 – t2)(1 + n2)(1 + s2)/{(mns + mnt – mst + nst + m – n + s + t)(mnst + mn + ms – mt + ns – nt – st – 1)}. (4.9) It is not difficult now to show that the common chord of T and T1 passes through E and F, where E has co-ordinates ((1 – t2/(1 + t2), 2s/(1 + s2)) and F has co-ordinates ((1 – m2)/(1 + m2), 2n/(1 + n2)) and the common chord has equation (1 + m2)(1 + t2)(1 – sn)(s – n)x – (m2 – t2)(1 + n2)(1 + s2)y + m2(n2s(1 + t2) + n(1 – t2)(1 + s2) + s(1 + t2)) – n2s(1 + t2) + n(1 – t2)(1 + s2) – s(1 + t2) = 0 (4.10) 5. The circle T2 The lines involved have the following equations AC': (1 + mt)x + (m – t)y = 1 – mt, BD': (n – s)x + (1 + ns)y = n + s, CA': (1 + mt)x + (t – m)y = 1 – mt, DB': (n – s)x – (1 +ns)y = – (n + s).
(5.1) (5.2) (5.3) (5.4)
Points involved have co-ordinates (x, y) P2= AC'^BD' x = – (mnst + mn + ms + mt – ns – nt – st – 1)/(mnst – mn + ms + mt + ns + nt – st + 1), y = 2(mnt + s)/(mnst – mn + ms + mt + ns + nt – st + 1);
(5.5)
Q2 = BD'^CA' x = – (mnst – mn – ms + mt + nt – ns + st – 1)/(mnst + mn – ms + mt + ns – nt + st + 1), y = 2(mnt + s)/(mnst + mn – ms + mt + ns – nt + st + 1);
(5.6)
R2= CA'^DB' x = – (mnst – mn – ms + mt + nt – ns + st – 1)/(mnst – mn + ms + mt + ns + nt – st + 1), y = 2(mst + n)/ (mnst – mn + ms + mt + ns + nt – st + 1);
(5.7)
S2 = DB'^CA' x = – (mnst +mn + ms + mt – ns – nt – st – 1)/(mnst + mn – ms + mt + ns – nt + st + 1), y = 2(mst + n)/(mnst + mn – ms + mt + ns – nt + st + 1).
(5.8)
It may now be checked that P2, Q2, R2, S2 lie on a circle T2, with centre O2 having co-ordinates (u, v), where u = (1 + n2)(1 + s2)(1 – m2t2)/{(mnst + mn – ms + mt + ns – nt + st + 1)( mnst – mn + ms 4
+ mt + ns + nt – st + 1)}, v = (1 + m2)(1 + t2)(1 + ns)(n + s)/{(mnst + mn – ms + mt + ns – nt + st + 1)( mnst – mn + ms + mt + ns + nt – st + 1)}. (5.9) It may also now be checked that O O0 and O O2 are the same line with equation (1 + m2)(1 + t2)(1 + ns)(n + s)x = (1 + n2)(1 + s2)(1 – m2t2)y
(5.10)
And that thus line passes through G, the midpoint of EF. It may also be shown, though we do not present details of the calculation, that circles T, T0, T2 are coaxal with limiting point G. 6. The circle T3 The lines involved have the following equations AD': (n + t)x + (1 – nt)y = n – t, BA': (1 + ms)x + (s – m)y = 1 – ms, CB': (1 – nt)x + (n + t)y = 1 + nt, DC': (s – m)x + (1 + ms)y = s + m. Points involved have co-ordinates (x, y) P3 = AD'^BA' x = – (mnst + mn – ms – mt – ns – nt + st +1)/(mnst – mn – ms – mt + ns + nt + st – 1), y = 2(t – mns)/(mnst – mn – ms – mt + ns + nt + st – 1); Q3 = BA'^CB' x = – (mns – mnt + mst + nst – m – n + s – t)/(mns – mnt + mst + nst + m + n – s + t), y = 2(ms + nt)/(mns – mnt + mst + nst + m + n – s + t); R3 = CB'^DC' x = – (mnst – mn + ms – mt + nt – ns – st + 1)/(mnst – mn – ms – mt + ns + nt + st – 1), y = 2(nst – m)/(mnst – mn – ms – mt + ns + nt + st – 1); S3 = DC'^ AD' x = (mns + mnt – mst + nst – m + n – s – t)/(mns – mnt + mst + nst + m + n – s + t), y = 2(mn + st)/(mns – mnt + mst + nst + m + n – s + t ).
(6.1) (6.2) (6.3) (6.4)
(6.5)
(6.6)
(6.7)
(6.8)
It may now be shown that P3, Q3, R3, S3 lie on the circle T3, centre O3 co-ordinates (u, v) where u = (1 + m2)(1 + t2)(s – n)(1 – sn)/{(mns – mnt + mst + nst + m + n – s + t)(mnst – mn – ms – mt + ns + nt + st – 1), v = (t2 – m2)(1 + n2)(1 + s2)/{( mns – mnt + mst + nst + m + n – s + t)(mnst – mn – ms – mt + ns + nt + st – 1)}. (6.9) 5
It may now be shown that O3 lies on OO1 and has common chord EF, showing that T, T1, T3 are coaxal.
Flat 4 Terrill Court 12-14 Apsley Road. BRISTOL BS8 2SP.
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Article 70 What happens when a Triangle is Rotated about its Orthocentre Christopher J Bradley
M
A
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E R B' L
1. Introduction The figure above shows a triangle ABC and its image A'B'C' after it has been rotated by 90o about its orthocentre H. It is as well to say immediately that the same results occur when the 1
angle differs from 90o and also when other rotation centres are chosen, such as the circumcentre, the incentre and the excentres. It appears possible that the results hold true when the rotation centre lies on a cubic, such as McCay’s cubic or the Orthocubic, but I am not able to draw cubics to test this and nor do accounts of these cubics indicate whether this property holds. I have to confess that calculations could have been carried out for an arbitrary angle, but the working would have been more difficult without adding any further significant interest. The main result is that circles AHA' and BHB' meet again at a point Z, which is also the intersection of AB and A'B'; the circles BHB' and CHC' meet again at a point X, which is also the intersection of BC and B'C'; and the circles CHC' and AHA' meet again at a point Y, which is also the intersection of CA and C'A'. A second result is that a conic can be drawn through the six points A, B, C, A', B', C'. If the lines AA', BB', CC' form the triangle LMN (with L = BB'^CC' etc.) then triangle LMN is similar to triangle ABC Also if D, E, F are the centres of the circles AHA', BHB', CHC' respectively, then circle DEF passes through X, Y and Z. Triangle DEF is also similar to triangle ABC. In what follows almost all of these results are established using Cartesian co-ordinates with the origin at H and HA along the line y = 0. 2. The points A, B, C, A', B', C' and the conic through these points Provided one is not concerned about the size of a triangle, then two parameters are sufficient to describe it. We choose the orthocentre H to be the origin and the axes along AH and parallel to BC through H. Co-ordinates of A, B, C may then be chosen to be: A(– 1 + vw, 0), B(vw, – v), C(vw, w). Since the rotation taking ABC to A'B'C' is 90o, the co-ordinates of A', B', C' are found to be A'(0, – 1 + vw), B'(v, vw), C'(– w, vw). It can now be checked that all six points lie on the conic with equation vw(v – 1)(w + 1)x2 + vw(v + 1)(w – 1)y2 + (3v2w2 – v2 – w2 – 2vw +1)x + vw(1 – vw)(3vw – v + w – 1)y + vw(1 – vw)(3vw + v – w – 1)x + 2v2w2(1 – vw)2 = 0. (2.1) 3. The nine lines and points L, M, N There are nine key lines in the configuration, the sides of the two triangles and the lines AA', BB', CC'. Their equations are as follows: AB: y + vx = v(vw – 1), (3.1) BC: x = vw, (3.2) CA: wx – y = w(vw – 1), (3.3) A'B': vy – x = v(vw – 1), (3.4) 2
B'C': C'A': AA' BB': CC':
y = vw, x + wy = w(vw – 1), x + y = 1 – vw, (1 + w)x – (1 – w)y = v(1 + w2), (1 + v)y – (1 – v)x = w(1 + v2).
We can now calculate the co-ordinates (x, y) of points L, M, N. L = BB'^CC' (x, y) = (1/2)(v(1 + w) + (1 – w), v(w – 1) + (1 + w)), M = CC'^AA' (x, y) = (1/2)(v(w – 1) – (1 + w), v(1 + w) – (1 – w)), N = AA'^BB' (x, y) = (1/2)(v(1 + w) – (1 – w),v(w – 1) – (1 + w)).
(3.5) (3.6) (3.7) (3.8) (3.9)
(3.10) (3.11) (3.12)
Verifying that the sides of triangle LMN are in proportion to the sides of ABC is left to the reader. CABRI II plus confirms with near certainty that this result holds. 4. Circles AHA', BHB', CHC' and the points D, E, F, X, Y, Z Circle AHA' passes through the origin and has centre (g, f), where g = (vw – 1)/2, f = (vw – 1)/2. Circle BHB' passes through the origin and has centre (g, f), where g = v(w + 1)/2, f = v(w – 1)/2. Circle CHC' passes through the origin and has centre (g, f), where g = w(v – 1)/2, f = w(v + 1)/2. These centres are respectively the points D, E, F. We can now obtain the co-ordinates (x, y) of the points X, Y, Z. X = BHB'^CHC':
(x, y) = (vw, vw). We may now check that X lies on both BC and B'C'.
Y = CHC'^AHA': (x, y) = {w(vw – 1)/(1 + w2)}(w + 1, w – 1). We may now check that Y lies on both CA and C'A'. Z = AHA'^BHB': (x, y) = {v(vw – 1)/(1 + v2)}(v – 1, v + 1). We may now check that Z lies on both AB and A'B'. 5. The circle DEFXYZ It may now be shown that the six points D, E, F, X, Y, Z lie on the circle with equation 2(x2 + y2) – (3vw + v – w – 1)x – (3vw – v + w – 1)y + 2vw(vw – 1) = 0. 6. Circles BHC, CHA, AHB and their centres R, S, T 3
(5.1)
Circle BHC passes through the origin and has centre R(g, f), where g = (1 + vw)/2, f = (w – v)/2. It may be verified that this circle contains L. Circle CHA passes through the origin and has centre S(g, f), where g = (vw – 1)/2, f = (v + w)/2. It may now be verified that this circle contains M. Circle AHB passes through the origin and has centre T(g, f), where g = (vw – 1)/2, f = – (v + w)/2. 7. The lines ST, TR, RS The equations of the lines ST, TR, RS are as follows: ST: TR: RS:
2x = vw – 1. 2wx – 2y = v(1 + w2). 2vx + 2y = w(1 + v2).
(7.1) (7.2) (7.3)
It happens in this configuration, but not in others that D = ST^AA', E = TR^BB', F = RS^CC'. Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article 71 More Special Cyclic Quadrilaterals Christopher J Bradley
C' S2 X P2
000000 °
D
A
R0
S0 D' E O2
O'
O0 C
O
R2
A' Q2
Q0
B'
Y
P0 B
1. Introduction This article is concerned with a cyclic quadrilateral ABCD in which AC is perpendicular to BD and its image A'B'C'D' after a rotation of 90 o about E = AC^BD. If AA'^BB' = P0, BB'^CC' = Q0, CC'^DD' = R0, DD'^AA' = S0, then P0Q0R0S0 is a rectangle and its circumcircle is coaxal with circles ABCD and A'B'C'D', all three meeting at points X, Y, where line XY passes through E. It is also the case that if AC'^BD' = P2, BD'^CA' = Q2, 1
CA'^DB' = R2, DB'^AC'= S2, then P2, Q2, R2, S2 are concyclic (but the circle is not part of the coaxal system). The centre O of ABCD, the centre O' of A'B'C'D' and E form three vertices of a square, with the centre O0 of P0Q0 R0S0 lying at the midpoints of the squares’ diagonals. In the following sections these results are proved using Cartesian co-ordinates with E as origin. 2. The circles ABCD and A'B'C'D' In choosing the co-ordinates of A, B, C, D we use the fact that EA.EC = EB.ED so with E as origin and EA along the positive direction of the x-axis we may set A(ab, 0), B(0, ac), C(– cd, 0), D(0, – bd), where a, b, c, d are positive real numbers. Since A', B', C', D' are the images of A, B, C, D under a 90 o rotation centred on E we obtain for their co-ordinates A'(0, ab), B'(– ac, 0), C'(0, – cd), D'(bd, 0). It is straightforward to find the equations of the circles ABCD and A'B'C'D' and these are ABCD: x2 + y2 + (cd – ab)x + (bd – ac)y – abcd = 0. (2.1) A'B'C'D' x2 + y2 + (ac – bd)x + (cd – ab)y – abcd = 0. (2.2) These two circles have the same radius equal to (1/2)√{(a2 + d2) (b2 + c2)}, the centre O of circle ABCD has co-ordinates ((ab – cd)/2, (ac – bd)/2) and the centre O' of circle A'B'C'D' has co-ordinates ((bd – ac)/2, (ab – cd)/2). 3. The lines AA', BB', CC', DD' and the circle P0Q0R0S0 The lines AA', BB', CC', DD' have the following equations AA': x + y = ab, BB': y – x = ac, CC': x + y + cd = 0, DD': x – y = bd.
(3.1) (3.2) (3.3) (3.4)
Clearly these lines form a rectangle. The co-ordinates of P0, Q0, R0, S0 are as follows: AA'^BB' = P0: (a(b – c)/2, a(b + c)/2). (3.5) BB'^CC' = Q0: (– c(a + d)/2, c(a – d)/2). (3.6) CC'^DD' = R0: (d(b – c)/2, – d(b + c)/2). (3.7) DD'^AA' = S0: (b(a + d)/2, b(a – d)/2). (3.8) The circle P0Q0R0S0 has equation 2x2 + 2y2 + (a + d)(c – b)x + (d – a)(b + c)y – 2abcd = 0.
(3.9)
The square of its radius is (1/8){(a2 + d2)(b2 + c2) + 4abcd} and its centre O0 has coordinates ((a + d)(b – c)/4, (a – d)(b + c)/4). It may now be checked that O0 is the mid point of OO'. In fact OO' has equation 2(a + d)(b – c)x + 2(a – d)(b + c)y – (a2 + d2)(b2 + c2) + 4abcd = 0. (3.10) The line perpendicular to this through E has equation 2
(a + d)(b – c)y – (a – d)(b + c)x = 0 and it may checked that O0 lies on this line.
(3.11)
4. The points X and Y We do not record the co-ordinates of both X and Y as it is the case that O0 is the midpoint of XY and the co-ordinates involved are lengthy expressions. X is defined as the point between A and D where EO0 meets circle ABCD and its coordinates are (x, y) where x = (a + d)(b – c)((a – d)√((a2 + d2)(b2 + c2) + 4abcd)√((a2 + d2)(b2 + c2) – 4abcd) + (a3 – d3) (b2 + c2) – ad(a – d)(b2 + 4bc + c2))/{4(a – d)((a2 + d2)(b2 + c2) – 4abcd)}, y = (b + c)((a – d)√((a2 + d2)(b2 + c2) + 4abcd)√((a2 + d2)(b2 + c2) – 4abcd) + (a3 – d3) (b2 + c2) – ad(a – d)(b2 + 4bc + c2))/{4((a2 + d2)(b2 + c2) – 4abcd)}.
(4.1)
It may now be checked that X and Y lie also on the circles centre O' and O0, and hence that these three circles form part of an intersecting coaxal system. 5. The lines AC', BD', CA', DB' and the circle P2Q2R2S2 The equations of the four lines are AC': cdx – aby = abcd. BD': acx + bdy = abcd. CA': cdy – abx = abcd. DB': – bdx – acy = abcd.
(5.1) (5.2) (5.3) (5.4)
The co-ordinates of the four points are as follows: P2 = AC'^BD' {ad/(a2 + d2)}(b(a + d), c(d – a)). Q2 = BD'^CA' {bc/(b2 + c2)}(d(c – b), a(b + c)). R2 = CA'^DB' {ad/(a2 + d2)}(–c(a + d), b(d – a)). S2 = DB'^ AC' {bc/(b2 + c2)}(a(c – b), – d(b + c)). It may now be shown that these four points lie on a circle with equation (a2 + d2)(b2 + c2)(x2 + y2) + (b – c)(bc(a3 + d3) –ad(a + d)(b2 + bc + c2))x – (b + c)(bc(a3 – d3) – ad(a – d)(b2 – bc + b2))y – (a2 + d2)(b2 + c2)abcd = 0.
(5.5) (5.6) (5.7) (5.8)
(5.9)
The centre and radius of this circle are not worth recording, except to say that the centre does not lie on OO'.
Flat 4 Terrill Court, 12-14 Apsley Road BRISTOL BS8 2SP
3
Article 72 What happens when you reflect a Triangle in any Line Christopher J Bradley
C'
B'
O'
D' Simson line of D
P A' L
A
D O
B
C
1. Introduction
1
In the figure a triangle ABC and its circumcircle are shown together with a reflection line L and the image triangle A'B'C'. What happens is that the line through A' parallel to BC, the line through B' parallel to CA and the line through C' parallel to AB are concurrent at a point P that always lies on circle A'B'C'. This result is proved below using Cartesian co-ordinates. CABRI II plus indicates that if another point D is taken on circle ABC and its Simson line is drawn then PD' is perpendicular to the Simson line of D, where D' is the reflection of D in L. If D' coincides with P, then the tangent to circle A'B'C'at P is perpendicular to the Simson line of P. Further, if L is dragged parallel to its initial position then there are two positions of L in which P also lies on circle ABC. In what follows the analytic condition for this to happen is obtained and comments from David Monk [1] as to their geometrical significance are described. 2. Choice of co-ordinates and the circle ABC We take Cartesian co-ordinates with L the line y = 0 and points A, B, C with co-ordinates A(a, d), B(b, e), C(c, h). With these choices the equation of the circle ABC is {a(e – h) + b(h – d) + c(d – e)}(x2 + y2) – (a2(e – h) + b2(h – d) + c2(d – e) – (h – d)(d – e)(e – h))x + (ab(a – b) + bc(b – c) + ca(c – a) – a(e2 – h2) – b(h2 – d2) – c(d2 – e2))y + a2(ce – bh) + b2(ah – cd) + c2(bd – ae) + aeh(e – h) + bhd(h – d) + cde(d – e) = 0. (2.1) 3. The lines BC, CA, AB and the point P The equations of the sides of the triangle ABC are as follows: BC: (h – e)x + (b – c)y = bh – ce, CA: (d – h)x + (c – a)y = cd – ah, AB: (e – d)x + (a – b)y = ae – bd.
(3.1) (3.2) (3.3)
We choose the line L to have equation y = 0 so that the image triangle’s vertices have coordinates A'(a, – d), B'(b, – e), C'(c, – h). The line through A' parallel to BC has equation (h – e)x + (b – c)y = a(h – e) – d(b – c). The line through B' parallel to CA has equation (d – h)x + (c – a)y = b(d – h) – e(c – a). The line through C' parallel to AB has equation (e – d)x + (a – b)y = c(e – d) – h(a – b).
2
(3.4) (3.5) (3.6)
It may now be checked that these three lines concur at the point P with co-ordinates (x, y) where x = (1/(a(e – h) + b(h – d) + c(d – e)))(a2(e – h) + b2(h – d) + c2(d – e) + bc(e – h) + ca(h – d) + ab(d – e)), y = (1/(a(e – h) + b(h – d) + c(d – e)))(a(d – e – h)(e – h) + b(e – h – d)(h – d) + c(h – d – e)(d – e). (3.7) The circle A'B'C' has the same equation as that of ABC (Equation (2.1)) except that d, e, h must be replaced by – d, – e, – h respectively. And then it may be checked that P lies on circle A'B'C'. 4. When P lies on circle ABC If one moves L (y = 0) parallel to itself then there are two cases when P lies also on circle ABC. The first case is when P lies on L and then L is the common chord of circles ABC and A'B'C'. The analytic condition for this is a(d – e – h)(e – h) + b(e – h – d)(h – d) + c(h – d – e)(d – e) = 0. (4.1) when y = 0 is the common chord. The second case is when the line L passes through the centre O of circle ABC and then O' is the same point as O. The analytic condition for this is (ab(a – b) + bc(b – c) + ca(c – a) – a(e2 – h2) – b(h2 – d2) – c(d2 – e2)) = 0. (4.2) David Monk [1] made a geometrical explanation of what is going on. I give his explanation, which is as follows: Let L be the line in which one reflects: A to A' etc. Your first result, that the parallels to BC through A', to CA through B and to AB through C' meet in a point P, is very easy by coordinates; With L as x-axis A as (a,d), B(b,e), C(c,h) the concurrence of the parallels requires the vanishing of a 3x3 determinant which is immediate by adding rows. Now for the question of when P lies on the circumcircle; this is very interesting. As a lucky guess I found that this is so when L is a diameter of the circumcircle. One suspects that there may be some connection with the orthopole and indeed this turns out to be so: If W is the orthopole of L with respect to ABC, H the orthocentre and P* the point diametrically opposite to the point of concurrence P then W (which is on the nine-point circle) is the midpoint of HP*. But the diameters are not the only lines with the property. Let L* be the line parallel to L through P*; it also passes through P', the reflection of P in L. Then L* also has the property: indeed the point of concurrence is P' - on both the reflection line and the circle. All the above I established by complex numbers with ABC as unit circle. Reference 1. D. Monk, private communication. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article 73 Constructing Two Coaxal systems by Incidence and Reflection Christopher J Bradley
R1 D' Z S2
R3 To S1
O1
Y
P2
S3 A, A'
O3
R2
Q1
O'
B S1
To R2
S0
O2 Q3
E
C, C'
O0
P0
P1
R0
X O W B' P3
Q2 Q0
D
1. Introduction The cyclic quadrilateral ABCD, centre O, with diagonals AC and BD at right angles at E is reflected in the line AC to form the cyclic quadrilateral A' B'C'D', centre O'.
1
The quadrilateral P0Q0R0S0, centre O0 is then defined as follows: the lines l0 parallel to AB through A', m0 parallel to BC through B', n0 parallel to CD through C', and p0 parallel to DA through D' are drawn and then P0 = l0^m0, Q0 = m0^n0, R0 = n0^p0, S0 = p0^l0 and P0Q0R0S0 is shown to be cyclic. Next the quadrilateral P1Q1R1S1, centre O1 is defined as follows: the lines l1 parallel to BC through A', m1 parallel to CD through B', n1 parallel to DA through C', and p1 parallel to AB through D' are drawn and then P1 = l1^m1, Q1 = m1^n1, R1 = n1^p1, S1 = p1^l1 and P1Q1R1S1 is shown to be cyclic. Next the quadrilateral P2Q2R2S2, centre O2 is defined as follows: the lines l2 parallel to CD through A', m1 parallel to DA through B', n1 parallel to AB through C', and p1 parallel to BC through D' are drawn and then P2 = l2^m2, Q2 = m2^n2, R2 = n2^p2, S2 = p2^l2 and P2Q2R2S2 is shown to be cyclic. Next the quadrilateral P3Q3R3S3, centre O3 is defined as follows: the lines l3 parallel to DA through A', m3 parallel to AB through B', n3 parallel to BC through C', and p3 parallel to CD through D' are drawn and then P3 = l3^m3, Q3 = m3^n3, R3 = n3^p3, S3 = p3^l3 and P3Q3R3S3 is shown to be cyclic. Finally it is shown that circles centre O', O1, O2 form 3 circles in an intersecting coaxal system and that circles centre O, O0, O3 also form 3 circles in an intersecting coaxal system. Proofs of these results are described in the following sections, using Cartesian co-ordinates with origin E and with AC and DB along the x- and y- axes. It should be mentioned that the results hold for an arbitrary cyclic quadrilateral reflected in any line (Cabri II plus conjecture), but the algebra necessary to prove this is technically too severe. 2. Circles ABCD and A'B'C'D' With origin and axes as described in Section 1 the co-ordinates of A, B, C, D may be chosen as A(– bd, 0). B(0, ab), C(ac, 0), D(0, – cd). The equation of the circle ABCD is therefore x2 + y2 + (bd – ac)x + (cd – ab)y – abcd = 0 (2.1) 2 2) 2 Its centre O has co-ordinates ((ac – bd)/2, (ab – cd)/2) and its radius is (1/2)√((d + a (b + c2)). Reflecting ABCD in the axis AC produces the image quadrilateral A'B'C'D' with co-ordinates A'(– bd, 0), B'(0, – ab), C'(ac, 0), D'(0, cd). The equation of circle A'B'C'D' may now be calculated and is x2 + y2 + (bd – ac)x + (ab – cd)y – abcd = 0. (2.2)
2
Its centre O' has co-ordinates ((ac – bd)/2, (cd – ab)/2) ) and its radius is (1/2)√((d2 + a2)(b2 + c2)). The sides of ABCD have equations: AB: ax – dy + abd = 0, BC: bx + cy – abc = 0, CD: dx – ay – acd = 0, DA: cx + by + bcd = 0.
(2.3) (2.4) (2.5) (2.6)
3. The circle P1Q1R1S1 centre O1 The lines parallel to BC, CD, DA, AB through A', B', C', D' respectively are bx + cy = – b2d, dx – ay = a2b, cx + by = ac2, ax – dy = – cd2.
(3.1) (3.2) (3.3) (3.4)
Intersections in pairs of these lines, as described in Section 1, produce the points P 1, Q1, R1, S1 whose co-ordinates are: P1: (ab(ac – bd)/(ab + cd), – b2(d2 + a2)/(ab + cd)), (3.5) 2 2 2 Q1: (a (b + c )/(ac + bd), ac(cd – ab)/(ac + bd)), (3.6) 2 2 2 R1: (cd(ac – bd)/(ab + cd), c (d + a )/(ab + cd)), (3.7) 2 2 2 S1: (– d (b + c )/(ac + bd), bd(cd – ab)/(ac + bd)). (3.8) The equation of circle P1Q1R1S1 may now be calculated and is (ab + cd)(ac + bd)(x2 + y2) – (a3b3 – c3d3 + abcd(cd – ab) + 2bc(a3c – d3b))x – (a3c3 – b3d3 + abcd(bd – ac) + 2da(dc3 – ab3))y – abcd(ab + cd)(ac + bd) = 0. Its centre is O1 with co-ordinates x, y), where x = (a3b3 – c3d3 + abcd(cd – ab) + 2bc(a3c – d3b))/(2(ab + cd)(ac + bd)), y = (a3c3 – b3d3 + abcd(bd – ac) + 2da(dc3 – ab3))/(2(ab + cd)(ac + bd)).
(3.9)
(3.10) (3.11)
4. The circle P2Q2R2S2 centre O2 The lines parallel to CD, DA, AB, BC through A', B', C', D' respectively are dx – ay = – bd2, cx + by = – ab2 ax – dy = a2c, bx + cy = c2d.
3
(4.1) (4.2) (4.3) (4.4)
Intersections in pairs of these lines, as described in Section 1, produce the points P 2, Q2, whose co-ordinates are: P2: (– b2(d2 + a2)/(ac + bd), bd(cd – ab)/(ac + bd)), Q2: (ab(ac – bd)/(ab + cd), – a2(b2 + c2)/(ab + cd)), R2: (c2(d2 + a2)/(ac + bd), ac(cd – ab)/(ac + bd)), S2: (cd(ac – bd)/(ab + cd),d2(b2 + c2)/(ab + cd)). The equation of circle P2Q2R2S2 may now be calculated and is (ab + cd)(ac + bd)(x2 + y2) + (a3b3 – c3d3 + abcd(cd – ab) + 2da(b3d – c3a))x + (a3c3 – b3d3 + abcd(bd – ac) + 2bc(a3b – d3c))y – abcd(ab + cd)(ac + bd) = 0. Its centre is O2, co-ordinates (x, y), where x = (a3b3 – c3d3 + abcd(cd – ab) + 2da(b3d – c3a))/(2(ab + cd)(ac + bd)), y = (a3c3 – b3d3 + abcd(bd – ac) + 2bc(a3b – d3c))/(2(ab + cd)(ac + bd)).
R2, S2 (4.5) (4.6) (4.7) (4.8)
(4.9)
(4.10) (4.11)
It may now be checked that O' is the midpoint of O1O2 and that the three circles centres O1, O2, O' intersect at a pair of points W, Z and hence form 3 circles in a coaxal system. (The coordinates of Z, W are very complicated and lengthy and no good purpose would be served by recording them.) 5. The circle P0Q0R0S0 and its centre O0 The lines parallel to AB, BC, CD, DA through A', B', C', D' respectively are ax – dy = – abd, bx + cy = – abc, dx – ay = acd, cx + by = bcd.
(5.1) (5.2) (5.3) (5.4)
Intersections in pairs of these lines, as described in Section 1, produce the points P0, Q0, R0, S0 whose co-ordinates are: P0: (– 2abcd/(ac + bd), ab(bd – ac)/(ac + bd)), (5.5) Q0: (ac(cd – ab)/(ab + cd), – 2abcd/(ab + cd)), (5.6) R0: (2abcd/(ac + bd), cd(bd – ac)/(ac + bd)), (5.7) S0: (bd(cd – ab)/(ab + cd), 2abcd/(ab + cd). (5.8)
The equation of circle P0Q0R0S0 may now be calculated and is (ab + cd)(ac + bd)(x2 + y2) + (ab – cd)(ac – bd)2x + (ab – cd)(bc(a2 + d2) – ad(b2 + c2))y – abcd(ab + cd)(ac + bd) = 0. (5.9)
4
Its centre is O0, co-ordinates (x, y), where x = – (ab – cd)(ac – bd)2/(2(ab + cd)(ac + bd)), y = – (ab – cd)(bc(a2 + d2) – ad(b2 + c2))/(2 (ab + cd)(ac + bd)).
(5.10) (5.11)
6. The circle P3Q3R3S3 and it centre O3 The lines parallel to DA, AB, BC, CD, through A', B', C', D' respectively are: cx + by = – bcd, ax – dy = abd, bx + cy = abc, dx – ay = – acd.
(6.1) (6.2) (6.3) (6.4)
Intersections in pairs of these lines, as described in Section 1, produce the points P 3, Q3, R3, S3 whose co-ordinates are: P3: (bd(ab – cd)/(ab + cd), – 2abcd/(ab + cd)), (6.5) Q3: (2abcd/(ac + bd), ab(ac – bd)/(ac + bd)), (6.6) R3: (ac(ab – cd)(ab + cd), 2abcd/(ab + cd)), (6.7) S3: (– 2abcd/(ac + bd), cd(ac – bd)/(ac + bd)). (6.8) The equation of circle P3Q3R3S3 may now be calculated and is (ab + cd)(ac + bd)(x2 + y2) + (bd – ac)(bc(a2 + d2) – da(b2 + c2))x – (bd – ac)(bc(a2 + d2) – da(b2 + c2))y – abcd(ac + bd(ab + cd) = 0.
(6.9)
The centre O3 has co-ordinates (x, y), where x = – (bd – ac)(bc(a2 + d2) – da(b2 + c2))/(2(ab + cd)(ac + bd)), y = (bd – ac)(bc(a2 + d2) – da(b2 + c2))/(2(ab + cd)(ac + bd)),
(6.10) (6.11)
It may now be checked that E is the midpoint of O3O0 and that O lies on the line O3O0. Furthermore the three circles centres O, O0, O3 have 2 common points X and Y, which means these 3 circles are part of a coaxal system. (The co-ordinates of X, Y are very complicated and lengthy and no good purpose would be served by recording them.)
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article 74 The Simson Line Porism Christopher J Bradley
L
A
D
M'
B'
C' P
N
O
N' B
C M
L' A'
1. Introduction Let ABC be a triangle and let A'B'C' be its image under a 180o rotation about the circumcentre O. Now let D be any point on the circumcircle S. Let P be the intersection of the Simson lines of D with respect to the two triangles. Then it turns out that the locus of D as P moves round the circumcircle is an inconic Ω of both triangles. This being the case, the circumcircle is the outer conic and this locus is the inconic of a porism. We call this the Simson line porism and the figure shows four triangles in the porism. The 1
triangles in the porism fall into pairs that are related by a 180 o rotation about O. The centre of Ω is at O. It should be remarked that if you have a single triangle and take points on S that are diametrically situated then the Simson lines of the two points are at right angles. But the locus of the point of intersection of the two Simson lines is the nine-point circle and not an inconic. In what follows we establish the porism using Cartesian co-ordinates and S the unit circle. 2. Points A, B, C and the Simson line of D We take S to be the unit circle and the point A to have co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)). Whereas point A has parameter a, points B, C, D have parameters b, c, d.
(2.1)
Since A' is the image of A in a 180o rotation then its parameter a' = – 1/a and similarly for the parameters of B' and C'. The side AB has equation (1 – ab)x + (a + b)y = (1 + ab). The perpendicular to AB through D therefore has equation (1 + d2)((a + b)x – (1 – ab)y) = 2abd + (a + b)(1 – d2) + a + b – 2d.
(2.2)
(2.3)
The foot of the perpendicular from D on to AB therefore has co-ordinates (x, y), where x = –1/{(1 + a2)(1 + b2)(1 + d2)}(a2(b2d2 + (b2 + d2 – 2bd) – 1) – 2a(1 + bd)(b – d) + b2d2 – b2 – d2 + 2bd – 1), y = 2/{(1 + a2)(1 + b2)(1 + d2)(a2b (1 + bd) + a(b – d)2 + d(1 + bd)). (2.4) The foot of the perpendicular from D on to CA has the same co-ordinates, but with c replacing b. The line joining these two points is the Simson line of D with respect to ABC and has equation (1 + d2)(abc – abd – acd – bcd – a – b – c + d)x – (1 + d2)(abcd + ab + ac + bc – ad – bd – cd – 1)y + a(b(c(3d2 + 1) – d(d2 – 1)) + (1 – d2)(1 + cd)) + b(1 – d2)(1 + cd) + c(1 – d2) – d(d2 + 3) = 0. (2.5) 3. The Simson line of D with respect to triangle A'B'C' and the point P Take Equation (2.5) and replace a, b, c by – 1/a, – 1/b, – 1/c respectively and this gives the Simson line of D with respect to triangle A'B'C'. These two Simson lines meet at a point P with co-ordinates (x, y), where
2
x = k(a2(b2(c2d2 – c2 – d2 + 4cd + 1) + b(4dc2 – 2cd2 + 2c) + (1 + c2)(1 – d2)) + 2a(2b2c2d – b2cd2 + b2c + b(1 – d2 )(1 + c2) + c(1 – d2) – 2d) + b2(1 + c2)(1 – d2) – 2b(c(d2 – 1) + 2d) + (c2 – 1)(1 – d2) – 4cd), y = 2k(a2(bc(b + c)(1 – d2) – 2bcd(1 – bc)) + a((1 – d2)(b2c2 – 1) – 2ad(b + c)(1 + bc)) + 2d(1 – bc) + (b + c)(d2 – 1)) , where k = 1/{(1 + a2)(1 + b2)(1 + c2)(1 + d2)}. (3.1) In order to get the locus of P as D varies, it is necessary to find the equation in x and y that eliminates d from the two equations in (3.1). In the abstract this would seem too difficult, but a lucky chance occurs as described in the next section. 4. The locus of P as D varies is an inconic of triangles ABC and A'B'C' What happens is that if you insert the co-ordinates (3.1) into the equation of AB you find that there is only one such meeting point when d = – 1/c. Similarly the curve touches BC, CA, A'B', B'C', C'A' when d = – 1/a, – 1/b, c, a, b respectively. We thus have 6 points on the locus. It is now found that these 6 points lie on the conic with equation {(1 + a2)(1 + b2)(1 + c2)}{4(1 + a2b2c2)x2 – 4(abc – 1)(a + b + c)xy + (a2b2c2 + 4abc(a + b + c) +a2b2 + b2c2 + c2a2 + a2 + b2 + c2 + 4(ab + bc + ca) + 1)y2} – 4(1 + ab)2(1 + bc)2(1 + ca)2 = 0. (4.1) Then one can verify that the general point with co-ordinates (3.1) does actually lie on the inconic with Equation (4.1). What we now have are two triangles inscribed in their common circumcircle and touching an inconic. Thus a porism is defined. There are now countless triangles that have the same properties, and which also have their images included when rotated about O by 180o
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article 75 The Double Simson Line Circle Christopher J Bradley
A P
C'
B'
H' O
H Q C
B
A'
1. Introduction In the figure ABC is a triangle and A'B'C' is its image under a rotation by 180o about the circumcentre O. With P a point on the circumcircle the Double Simson lines of P with respect to the triangles are drawn, showing them to meet at a point Q. The locus of Q as P moves round the circumcircle turns out to be the circle centre O on HOH' as diameter. In what follows this 1
property is proved, using Cartesian co-ordinates. This article should be read in conjunction with Article 74, where the Cartesian frame is defined along with the co-ordinates of the vertices of the two triangles. 2. The reflections of P in the sides of ABC In Article 74 [1] A has co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2), with B, C, P having parameters b, c, d. The reflected points A', B', C' then have co-ordinates – 1/a, – 1/b, – 1/c respectively. Equation (2.4) of that Article gives the co-ordinates of the foot of the perpendicular of P on to AB. From there it is straightforward to work out the co-ordinates of the reflection of P in AB and these are x = {– 1/((1 + a2)(1 + b2)(1 + d2))}(a2(b2(d2 + 3) – 4bd + d2 – 1) + 4a(d – b)(1 + bd) + b2(d2 – 1) + 4bd – 3d2 – 1), y = {1/((1 + a2)(1 + b2)(1 + d2))}(2(a2(b2d + 2b – d) + 2a(b – d)2 – d(b2 – 2bd – 1))). (2.1) The co-ordinates of the reflection of P in sides BC, CA may now be written down by appropriate changes of parameter. The line passing through all these points, which is what we call the Double Simson line of P with respect to ABC, has equation (abc – abd – bcd – cad – a – b – c + d)x – (abcd + ab + bc + ca – ad – bd – cd – 1)y + (3abc – abd – bcd – cad + a + b + c – 3d) = 0. (2.2) The equation of the Double Simson line of P with respect to triangle A'B'C' may be witten down from Equation (2.2) by replacing a, b, c by – 1/a, – 1/b, – 1/c respectively. 3. The point Q where the Double Simson lines meet The point Q where the two Double Simson lines meet has very complicated co-ordinates (x, y) where k = 1/((1+ a2)(1 + b2)(1 + c2)(1 + d2)) and x = k(a2(b2(3c2(d2 – 1) + 8cd – d2 + 1) + 4bc(2cd – d2 + 1) + (1 – d2)(1 + c2)) + 4a(b2c(2cd – d2 + 1) + b(1 – d2)(1 + c2)+ c(1 – d2) – 2d) + b2(1 – d2)(1 + c2) – 4b(c(d2 – 1) + 2d) + c2(1 – d2) – 8cd + 3d2 – 3), (3.1) y = 2k(a2(b2(3c2d + 2c(1 – d2) – d) – 2bc(c(d2 – 1) + 2d) – d(1 + c2)) – 2a(b2c(c(d2 – 1) + 2d) + 2bd(1 + c2) + 2cd – d2 + 1) – b2d(1 + c2) – 2b(2cd – d2 + 1) – c2d + 2c(d2 – 1) + 3d). (3.2) It may now be verified that the locus of Q as P varies is the circle centre O and radius squared equal to 9a2b2c2 + 8abc(a + b + c) + a2b2 + b2c2 + c2a2 + 8(ab + bc + ca) + a2 + b2 + c2 + 9. (3.3) This circle has HOH' as diameter.
2
Reference 1. C.J.Bradley, Article 74, The Simson Line Porism, (2010).
Flat 4, Terrill Court, 12-14, Apsley Road BRISTOL BS8 2SP.
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Article 76 The Double Simson Line Conic Christopher J Bradley
M'
Q
N A
P N' R
M
O
B
L
C
L'
1. Introduction In the Figure ABC is a triangle and P is a point on its circumcircle. Transversal LMN is the Double Simson line formed by points that are reflections of P in the sides of ABC. This always 1
passes through the orthocentre H. Points L', M', N' are the harmonic conjugates of L, M, N respectively with respect to points B, C and C, A and A, B. The three Cevians AL', BM', CN', are concurrent at Q. It is shown in what follows that the locus of Q is a circumconic of ABC and whose equation is related to H in a particularly simple way. Areal co-ordinates are used throughout with ABC as triangle of reference. 2. The Equation of the Double Simson Line It is well-known (see Bradley [1] p 422) that the reflection of the point with co-ordinates (d, e, f) in the side BC has co-ordinates (– d, (a2(d + e) + d(b2 – c2))/,a2, (a2(d + f) – d(b2 – c2))/a2). If now the point P that is reflected lies on the circumcircle and has co-ordinates (a2t(1 – t), – b2(1 – t), – c2t) then the reflected point U in BC has co-ordinates (x, y, z), where x = a2t(t – 1), y = (1 – t)(a2t + b2(t – 1) – c2t), z = – t(a2(t – 1) + b2(1 – t) + c2t). (2.1) Similarly when P is reflected in CA the reflected point V has co-ordinates (x, y, z), where x = (1 – t)(a2(t – 1) – b2 + c2), y = b2(1 – t), z = a2(1 – t) + b2(t – 1) – c2. (2.2) The line UV is the Double Simson line with equation (a2 – b2 – c2)x + t(a2 – b2 + c2)y + (1 – t)(a2 + b2 – c2)z = 0.
(2.3)
3. The points L, M, N These are the points where the Double Simson line meets the sides of ABC. Their co-ordinates are: L: (0, (a2 + b2 – c2)(t – 1), (a2 – b2 + c2)t), M: ((a2 + b2 – c2)(t – 1), 0, a2 – b2 – c2), (3.1) 2 2 2 2 2 2 N: ((a – b + c )t, (b + c – a ), 0). The points L', M', N' are the harmonic conjugates of L, M, N respectively and their co-ordinates may be obtained from those of L, M, N by changing the sign of one of their non-zero coordinates. 4. The associated Cevian point Q and the locus of Q The lines AL', BM', CN' meet at the point Q with co-ordinates (x, y, z), where x = t(1 – t)/(b2 + c2 – a2), y = – (1 – t)/(c2 + a2 – b2), z = – t(a2 + b2 – c2).
2
(4.1)
The locus of Q as P moves around the circumcircle and as the Double Simson line rotates about H is evidently yz/(b2 + c2 – a2) + zx/(c2 + a2 – b2) + xy/(a2 + b2 – c2) = 0. (4.2) The form of this equation is noteworthy, since this locus is a circumconic of ABC in which the coefficients of yz, zx, xy are respectively the x-, y-, z- components of the orthocentre H. This leads us to ask the question: Suppose we had started with the lines all passing through a point with co-ordinates (l, m, n), will the locus of Q be lyz + mzx + nxy = 0? The answer is ‘Yes’. The working is simply a repeat of the above analysis in the new context. In particular if the lines pass through the symmedian point K(a2, b2, c2) then the locus of Q is the circumcircle.
Reference 1. C.J.Bradley, The Algebra of Geometry, Highperception, Bath, (2007).
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Article 77 The Circle centre O and radius OH Christopher J Bradley
A
Z
Q
X
DeL
O P H
B
R
Y
C
T
1. Introduction 1
The figure above illustrates a number of related results about the circle Ω centre the circumcentre O of triangle ABC and radius OH, where H is the orthocentre of triangle ABC. Points on this circle are defined as follows: X, Y, Z are the intersections of AH, BH, CH respectively with Ω. Points P, Q, R are defined so that XOP, YOQ, ZOR respectively are diameters of Ω. The following results now hold for the fourteen circles in the figure: (i) Circles BXC, CYA, AZB are concurrent at deLongchamps point deL, which is at the other end to H of the diameter HO of Ω; (ii) Circles AYZ, BZX, CXY are concurrent at a point T on the circumcircle of ABC; also the ratios TC/TB and deL Z/deL Y are equal; (iii)Circles BPC, CQA, ARB are concurrent at H; (iv) Circles AQR, BRP, CPQ are concurrent at T. Work in the following sections is carried out using Cartesian co-ordinates with A, B, C lying on the rectangular hyperbola xy = 1. 2. Points X, Y, Z, deL and the Circles passing through deL We take A, B, C to have co-ordinates (a, 1/a), (b, 1/b), (c, 1/c) respectively. It may then be shown that the point O equidistant from these three points has co-ordinates (x, y), where x = {abc(a + b + c) + 1}/(2abc), y = {a2b2c2 + bc + ca + ab}/(2abc). (2.1) The equation of the circle centre O and radius OH may now be determined and it is a2b2c2(x2 + y2) – abc(abc(a + b + c) + 1)x – abc(a2b2c2 + bc + ca + ab)y – 2a4b4c4 – a2b2c2(ab + bc + ca) – abc(a + b + c) – 2 = 0.
(2.2)
The co-ordinates of deL are (x, y), where x = (abc(a + b + c) + 2)/(abc),
(2.3)
y = (2a2b2c2 + ab + bc + ca)/(abc).
The co-ordinates of X are (x, y), where x = (3a2b3c3 + abc(a + 2b + 2c) + 2)/abc( 1 + b2c2), y = (2a2b3c3 + 2abc(b + c) + b2c2 + 3)/(a(1 + b2c2)).
(2.4)
It may now be shown by the determinantal method (with columns x 2 + y2, x, y, 1 of the coordinates of the four points concerned that B, C, X, deL are concyclic. The co-ordinates of Y and Z may be obtained from Equation (2.4) by cyclic change of a, b, c, so that as B, C, X, deL are concyclic then so are C, A, Y, deL and A, B, Z, deL. 3. Points P, Q, R, T and Circles ABC, AYZ, BZX, CXY 2
The point P is at the other end of the diameter XO of the circle centre O passing through H, X, Y, Z. Its co-ordinates are (x, y), where x = {–1/(abc(1 + b2c2)}(2a2b3c3 + abc(b + c)(1 – b2c2) + (1 – b2c2)), y = {–1/(abc(1 + b2c2)}(a2b2c2(b2c2 – 1) + a(b + c)(b2c2 – 1) + 2bc). (3.1) The co-ordinates of P and Q may be obtained from Equation (3.1) by cyclic change of a, b, c. The co-ordinates of O are given by Equation (2.1) and those of H are (– 1/abc, – abc) so the equation of circle ABC is abc(x2 + y2) – (abc(a + b + c) + 1)x – (a2b2c2 + ab + bc + ca)y + (abc(ab + bc + ca) + (a + b + c)) = 0. (3.2) The co-ordinates (x, y) of a special point T on the circumcircle ABC are x = (1/k)(a6b3c3(1 + 3b2c2) + a5b3c3(3b3c2 + 3b2c3 + b + c) + a4bc(b4c2 + 18b3c3 + b2(c4 – 1) + 4bc – c2) + a3bc(b5c2 + b4c3 + b3(c4 – 1) + b2c(c4 + 7) + 7bc2 – c3) + a2(4b4c2 + 7b3c3 + 2b2(2c4 – 1) + 5bc – 2c2) – abc(b3c2 + b2c3 – 5b – 5c) – 2b2c2), y = (– 1/k)(2a6b4c4 + a5b2c2(b + c)(1 – 5b2c2) + a4b2c2(2b4c2 – 5b3c3 + 2b2(c4 – 2) – 7bc – 4c2) + a3(b + c)(b4c2 – 8b3c3 + b2(c4 – 1) – c2) + a2bc(b4c2 – 4b3c3 + b2(c4 – 1) – 18bc – c2) – a(b + c)(b2c2 + 3) – bc(b2c2 + 3)), (3.3) where k = abc(a4b2c2(9b2c2 + 1) + 8a3b2c2(b + c) + a2(b4c2 + 8b3c3 + b2(c4 + 1) + 8bc + c2) + 8abc(b + c) + b2c2 + 9). (3.4) (The author actually obtained T as the intersection of the common chord of circles ABC and AYZ with circle ABC, but many of the expressions were technically very complicated and lengthy and hence difficult to record accurately.) It may now be shown by a determinantal method that circles AYZ, BZX, CXY all pass through T. 4. Circle BCH passes through P, circle CAH passes through Q and circle ABH passes through R The equation of circle BCH is comparatively easy to obtain and is abc(x2 + y2) + (abc(a – b – c) +1)x + (a2b2c2 – ab – ca + bc)y – abc(ab + ca – bc) + a – b – c = 0. (4.1)
3
The co-ordinates of P are given in Equation (3.1) and it may now be shown by substitution that P lies on circle BCH. Similarly Q lies on circle CAH and R lies on circle ABH. 5. Circles AQR, BRP, CPQ pass through T The determinantal method can once again be employed to show these three circles pass through T. There are now 7 circles in all passing through T, so it must be a significant point as far as this construction is concerned. In fact it may be shown that its angular disposition with respect to B and C is the same as the angular disposition of deL with respect to Y and Z on circle XYZH. It will be recalled that triangle XYZ is indirectly similar to triangle ABC with respect to an axis through O that bisects angle COZ (and BOY and AOX), being a reflection in that axis and an enlargement/reduction through O. See Article 1,[1]. In fact deL and T are related by the same operations. Proof is left to the reader. Flat 4 Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article 78 More on the Seven-Point Circle Christopher J Bradley
A
Z
V
O X
T U
K Y
W C
B
P
1. Introduction The Seven-Point Circle of triangle ABC passes through its Symmedian Point K and the lines AK, BK, CK meet the circle again at X, Y, Z respectively. These are the vertices of the Second Brocard Triangle. The circles BKC, CKA, AKB meet the Seven-Point Circle again at U, V, W respectively. It is shown that UX, VY, WZ are concurrent at a point T. (This type of involution
1
holds for any circle passing through a given point not lying on the sides of ABC, see Article 3 of Bradley [1]. It is also shown that circles BXC, CYA, AZB all pass through the circumcentre O. We prove that circles AYZ, BZX, CXY, AVW, BWU, CUV all pass through a point P on the circumcircle of ABC. In what follows areal co-ordinates are used with ABC as triangle of reference. 2. Points X, Y, Z and circles BXC, CYA, AZB The equation of the Seven-Point Circle, see Bradley [2], is b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0.
(2.1)
The equation of AK is c2y = b2z and this meets the Seven-Point circle at the point X with coordinates X(b2 + c2 – a2, b2, c2).Similarly points Y and Z have co-ordinates Y(a2, c2 + a2 – b2, c2) and Z(a2, b2, a2 + b2 – c2). The equation of circle BXC is b2c2x2 – a2(b2 + c2 – a2)yz + b2(a2 – b2)zx + c2(a2 – c2)xy = 0.
(2.2)
It may now be checked that O(a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)) lies on this circle. Circles CYA, AZB have equations that may be written down from Equation (2.2) by cyclic change of a, b, c and x, y, z. These two circles also pass through O. 3. Circles AYZ, BZX, CXY and the point P We can now determine the equation of circle AYZ, which is c2a2(c2 – a2)y2 – a2b2(a2 – b2)z2 – a2(a2 – b2)(c2 – a2)yz + b2(a2b2 – b4 + b2c2 – c4)zx + c2(c2a2 – b4 + b 2c2 – c4)xy = 0.
(3.1)
This circle meets the circumcircle with equation a2yz + b2zx + c2xy = 0 at the point P with coordinates (x, y, z), where x = a2(a2 – b2)(c2 – a2), y = b2(b2 – c2)(a2 – b2), (3.2) 2 2 2 2 2 z = c (c – a )(b – c ). The symmetrical form of these expressions shows that P also lies on circles BZX and CXY. 4. Circles BKC, CKA, AKB and points U, V, W
2
The equation of circle BKC is 3b2c2x2 – a2(a2 + b2 + c2)yz – b2(a2 + b2 – 2c2)zx – c2(c2 + a2 – 2b2)xy = 0.
(4.1)
This circle meets the Seven-Point circle with equation (2.1) at the point U with co-ordinates U(a2(a2 + b2 + c2), b2(4c2 + a2 – 2b2), c2(4b2 + a2 – 2c2)). The co-ordinates of V and W may now be written down by cyclic change of x, y, z and a, b, c. The equation of UX may now be obtained and is 6b2c2(b2 – c2)x + 2c2(a4 + 2a2b2 – a2c2 – (b2 + c2)(2b2 – c2))y – 2b2(a4 + 2a2c2 – a2b2 + (b2 + c2)(b2 – 2c2))z = 0.
(4.2)
The equation of VY and WZ may now be written down using cyclic change of x, y, z and a, b, c. It may now be shown that all three lines pass through a point T with co-ordinates (x, y, z), where x = a2(2b2 + 2c2 – a2), y = b2(2c2 + 2a2 – b2), (4.3) 2 2 2 2 z = c (2a + 2b – c ). It is left to the reader to show that circles AVW, BWU and CUV all pass through P.
References 1. C.J. Bradley, Article 3. 2. C.J. Bradley, The Algebra of Geometry, Highperception, Bath (2007).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article 79 Conics in the Ex-circle Configuration Christopher J Bradley A'
T
V N To T F
A E D' J
Z
S
Y
Q K H
B To B'
E'
L
P
F' U
C
X
W
M To C' To R
D
C' B'
1. Introduction
R
Given a triangle ABC there are three ex-circles. The one opposite A has centre D and touches BC at X, CA at M and AB at W. The one opposite B has centre E and touches CA at Y, AB at N and BC at U. The one opposite C has centre F and touches AB at Z, BC at L and CA at V. 1
In what follows we show that a conic passes through L, M, N, U, V, W and that three further conics exist passing through (i) L, U, Z, Y, V, N, (ii) M, V, X, Z, W, L and (iii) N, W, Y, X, U, M. These conics involve the points of contact with the sides of pairs of ex-circles. Lines LW, MU meet at R; MU, NV meet at S; NV, LW meet at T. It follows that lines SE, TF and RD are concurrent at a point J. It also follows that lines RA, SB and TC are concurrent at H, the orthocentre of triangle ABC. It is then proved that the centre of conic LUZYVN is a point D' lying on EF, the centre of conic MVXZWL is a point E' lying on FD and the centre of conic NWYXUM is a point F' lying on DE. Furthermore DD', EE', FF' are concurrent at a point Q. Conics MVXZWL and NWYXUM obviously meet at M, X and W but their fourth point of intersection determines a point A', with B', C' similarly defined. It is proved that AA', BB', CC' meet at the symmedian point K of triangle ABC and also triangle A'B'C' is an enlargement of triangle ABC. Finally it is shown that the lines RA', SB', TC' are concurrent at a point P. Areal co-ordinates are used with ABC the triangle of reference. DERIVE and CABRI II plus were used in the drawing and analysis. 2. Revision of the co-ordinates of main points and the equations of the ex-circles The centres of the ex-circles are D(– a, b, c), E(a, – b, c), F(a, b, – c). Points of contact of the excircles with interior points of the sides of ABC are X(0, c + a – b, a + b – c), Y(b + c – a, 0, a + b – c), Z(b + c – a, c + a – b, 0). The other six points of contact of the ex-circles with exterior points of the sides of ABC are U(0, a – b – c, a + b + c), V(a + b + c, 0, b – c – a), W(c – a – b, a + b + c, 0), L(0, a + b + c, a – b – c), M(b – c – a, 0, a + b + c), N(a + b + c, c – a – b, 0). The ex-circle opposite A passing through W, X and M is (a + b + c)2x2 + (a + b – c)2y2 + (a – b + c)2z2 – 2(a2 – (b – c)2)yz + 2((c + a)2 – b2)zx + 2((a + b)2 – c2)xy = 0. (2.1) The ex-circles opposite B and C may be written down by cyclic change of x, y, z and a, b, c. 3. The four conics The equation of the conic LMNUVW through the external points of contact is (a2 – (b + c)2)(a2 – (b – c)2)x2 + (b2 – (c + a)2)(b2 – (c – a)2) y2 + (c2 – (a + b)2)(c2 – (a – b)2)z2 – 2(a2 + (b + c)2 ) (a2 – (b – c)2)yz – 2(b2 + (c + a)2)(b2 – (c – a)2)zx – 2(c2 + (a + b)2)(c2 – (a – b)2)xy = 0. (3.1) 2
The equation of the conic LUZYVN through the points of contact of the ex-circles opposite B and C is (a2 – (b – c)2)x2 + (a2 – (b + c)2)y2 + (a2 – (b + c)2)z2 – 2(a2 + (b + c)2)yz + 2(a2 + b2 – c2)zx + 2(a2 – b2 + c2)xy = 0. (3.2) The equations of the conics MVXZWL and NWYXUM may be written down from equation (3.2) by repeated cyclic change of x, y, z and a, b, c. 4. Points R, S, T and the concurrence of RD, SE, TF The equation of UM is (a2 – (b + c)2x – ((a + c)2 – b2)y + ((a – b)2 – c2)z = 0.
(4.1)
The equations of VN and WL may now be written down by repeated cyclic change of x, y, z and a, b, c. UM and VN meet at the point S whose co-ordinates are (x, y, z), where x = ((a + c)2 – b2)(a2 + b2 – c2), y = 2b(c + a)((c – a)2 – b2), z = ((c + a)2 – b2)(b2 + c2 – a2).
(4.2)
The equation of SE is b(a4 + 2a2(c2 – b2) + (b2 – c2)(b2 + 3c2))x +((c + a)2 – b2)(a3 + a2c – a(b2 + c2)+ c(b2 – c2))y + b(3a4 – 2a2(b2 + c2) – (c2 – b2)2)z = 0. (4.3) The equations of TF and RD may now be written down by repeated cyclic changes of x, y, z and a, b, c. The determinant whose rows are the coefficients of x, y, z in the equations of SE, TF, RD may now be shown to vanish, showing these lines are concurrent at a point J. The actual co-ordinates of J are technically very complicated and lengthy and are therefore not recorded here. 5. RA, SB, TC meet at H The equation of the line RA is (c2 + a2 – b2)y = (a2 + b2 – c2)z and RA clearly passes through H, the orthocentre of ABC, as do SB and TC. 6. The centres of the conics
3
The centre D' of the conic LUZYVN has co-ordinates (– a2, b(c – b), c(b – c)) and the centres E', F' of the conics MVXZWL and NWYXUM may be written down by repeated cyclic changes of x, y, z and a, b, c. It may now be shown that DD', EE', FF' are concurrent at a point Q with coordinates (a(3a – b – c), b(3b – c – a), c(3c – a – b)). 7. Where the conics meet The conics MVXZWL and NWYXUM meet at M, X, W and at a fourth point A' whose coordinates are (– (b + c)(ab + ca + b2 + c2), b2(a + b + c), c2(a + b + c)). Points B', C' may be similarly defined and clearly AA' passes through the symmedian point K of triangle ABC. The following facts are also easily checked: triangle A'B'C' is an enlargement of triangle ABC centre K, which is also the symmedian point of triangle A'B'C'. 8. Lines RA', SB', TC' are concurrent The equation of the line RA' is (b – c)(a2 – (b + c)2)x + (a2(b + c) + 2ca(c – b) – (b + c)(b2 – c2))y – (a2(b + c) + 2ab(b – c) + (b + c)(b2 – c2))z = 0.
(8.1)
The lines SB', TC' have equations that may be written down from equation (8.1) by repeated cyclic change of x, y, z and a, b, c. The determinantal method involving the coefficients of x, y, z in these equations show that RA', SB', TC' are concurrent at a point P, which has lengthy and technically involved expressions.
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Article 80 Incircle and Excircle Conics Christopher J Bradley
V
F
N A E Z
B' Y'
Z'
B
L
X'
E' F' C' Y P I K D' A' X C
U
W
M
D
Fig.1 Four Conics through points of contact of the Incircle and Excircles 1
1. Introduction In triangle ABC points X, Y, Z are the points of contact of the incircle with BC, CA, AB respectively. Points X', Y', Z' are the internal; points of contact on the sides BC, CA, AB respectively of the three excircles and the external points of contact are L, U with BC, M, V with CA and N, W with AB. It is shown in this short article that a conic S passes through X, X', Y, Y', Z, Z' and a conic S1 passes through X, Y, Z, X', W, M and a conic S2 passes through X, Y, Z, Y', U, N and a conic S3 passes through X, Y, Z, Z', V, L. Conics S2 and S3 meet at A', conics S3 and S1 meet at B' and conics S1 and S2 meet at C'. The centres of conics S1, S2, S3 are at D', E', F' respectively. It is then shown that AA', BB', CC' are concurrent at the symmedian point K, that AD', BE', CF' are concurrent at the incentre I, and that XD', YE', ZF' are concurrent at a point P. 2. The conic XX'YY'ZZ' Its equation is (a – b + c)(a + b – c)(a – b – c)x2 + (b – c + a)(b + c – a)(b – c – a)y2 + (c – a + b)(c + a – b)(c – a – b)z2 + 2(b + c – a)(a2 + (b – c)2)yz + 2(c + a – b)(b2 + (c – a)2)zx + 2(a + b – c)(c2 + (a – b)2)xy = 0. (2.1) This may be checked by showing that the six points X(0, a + b – c, a + c – b), Y(b + a – c, 0, b + c – a), z(c + a – b, c + b – a, 0), X'(0, c + a – b, a + b – c), Y'(b + c – a, 0, b + a – c), Z'(b + c – a, c + a – b, 0) all have co-ordinates satisfying Equation (2.1). 3. The conics S1, S2, S3 The points of contact of the excircles with the sides of the triangle are U(0, a – b – c, a + b + c), V(a + b + c, 0, b – c – a), W(c – a – b, a + b + c, 0), L(0, a + b + c, a – b – c), M((b – c – a, 0, a + b + c), N(a + b + c, c – a – b, 0). It may now be verified that the equation of S1 is (a + b + c)(a – b – c)x2 + (a + b – c)(a – b + c)y2 + (a + b – c)(a – b + c)z2 – 2(a2 + (b – c)2)yz + 2(a2 + b2 – c2)zx + 2(a2 – b2 + c2)xy = 0. (3.1) The equations of S2 and S3 may be written from Equation (3.1) by repeated cyclic changes of x, y, z and a, b, c. 4. The points A', B', C' and the concurrency of AA', BB', CC'
2
Conics S2 and S3 meet at the point labelled A' whose co-ordinates are ((b + c)(ab + ca – b2 – c2), b2(b + c – a), c2(b + c – a)). The co-ordinates of B' and C' follow by repeated cyclic change of x, y, z and a, b, c. It is clear that AA' passes through K(a2, b2, c2), as do BB' and CC'. 5. The centres D', E', F' of conics S1, S2, S3 The co-ordinates of D' are (a2(a2 +(b – c)2), b(b + c)(a2 + (b – c)2), c(b + c)(a2 + (b – c)2). The coordinates of E' and F' follow by repeated cyclic change of x, y, z and a, b, c. Clearly AD' passes through I(a, b, c) as do BE' and CF'. 6. The concurrency of XD', YE', ZF' The equation of XD' is (b + c – a)(b2 – c2)x +a2(c + a – b)y – a2(a + b – c)z = 0.
(6.1)
The equations of YE' and ZF' follow by repeated cyclic changes of x, y, z and a, b, c. It may now be verified that these three lines meet at a point P with co-ordinates (x, y, z), where x = a2(c + a – b)(a + b – c)(b2 + c2 – a2), y = b2(a + b – c)(b + c – a)(c2 + a2 – b2), (6.2) 2 2 2 2 z = c (b + c – a)(c + a – b)(a + b – c ).
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Article: CJB/2010/81
A Radical Centre that lies on OI Christopher J Bradley
A
F N
V T
E S
O
W
I J
B
M
R
L
U
C
D
Fig.1 The radical centre J lies on the line OI 1. Introduction Fig.1 shows the very familiar configuration in which, when ABC is a triangle with incentre I, the circles BIC, CIA, AIB are drawn. These circles have centres D, E, F respectively. It is known that AI, BI, CI pass through D, E, F and that D is the midpoint of the arc BC etc. DE cuts BC at U, CA at M and AB at Z, EF cuts CA at V, AB at N and BC at X and FD cuts AB at 1
W, BC at L and CA at Y. Since triangles ABC and DEF are in perspective, X, Y, Z are collinear. The three circles AWM, BUN, CVL are drawn and the purpose of this brief note is to show that the radical centre of these three circles is a point J lying on the line OI, where O is the circumcentre of ABC. J is actually the isogonal conjugate of Gergonne’s point. The equation of the conic UVWLMN is also obtained. We use areal co-ordinates, and in order to establish the results it is necessary to revise some well-known material. 2. Areal co-ordinates of main points and equations of important lines These are: D(– a2, b(b + c), c(b + c)), E(a(c + a),– b2, c(c + a)), F(a(a + b), b(a + b), – c2). The equation of EF is bcx – c(c + a)y – b(a + b)z = 0.
(2.1)
The equations of FD and DE may now be written down by repeated cyclic change of x, y, z and a, b, c. EF meets AB at N(c + a, b, 0), EF meets CA at V(a + b, 0, c). and EF meets BC at X(0, b(a + b), – c(c + a)). L has co-ordinates (0, a + b, c, W has co-ordinates (a, 0, b + c), Y has co-ordinates ( – a(a + b), 0, c(b + c)), M has co-ordinates (a, 0, b + c), U has co-ordinates(0, b, c + a) and Z has co-ordinates (a(c + a), – b(b + c), 0). The line XYZ has equation bc(b + c)x + ca(c + a)y + ab(a + b)z = 0.
(2.2)
The points U, V, W, L, M, N lie on a conic whose equation is bc(b + c)x2 + ca(c + a)y2 + ab(a + b)z2 – a(a2 + a(b + c) + 2bc)yz – b(b2 + b(c + a) + 2ca)zx – c(2 + c(a + b) + 2ab)xy = 0. (2.3) The equation of WM is (b + c)x = a(y + z),
(2.4)
and this is the line through I parallel to BC. Similarly UN is the line through I parallel to CA and VL is the line through I parallel to AB. 3. The three circles, their radical axes and the point J The equation of the circle AWM is a2yz + b2zx + c2xy – (ux + vy + wz)(x + y + z) = 0,
2
(3.1)
where u = 0, v = ac2/(a + b + c), w = ab2/(a + b + c). The equations of the circles BUN and CVL may now be written down by repeated cyclic change of x, y, z and a, b, c. It follows that the radical axis of circles AWM and BUN is bc2x – ac2y + ab(a – b)z = 0.
(3.2)
The equations of the other two radical axes may now be written down by repeated cyclic change of x, y, z and a, b, c. The three lines meet at the radical centre J with co-ordinates (x, y, z), where x = a2(b + c – a), y = b2(c + a – b), z = c2(a + b – c).
(3.3)
J is the isogonal conjugate of Gergonne’s point (1/(b + c – a), 1/(c + a – b), 1/(a + b – c)) and J lies on IO, as may be discovered by writing the co-ordinates of the three points I, J, O as rows of a determinant and showing that the determinant vanishes. J is actually X55 in Kimberling’s notation and the Encyclopedia of Triangle centres wrongly says X1, X3, and X35 are collinear.
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Article: CJB/82/2010
Cevian derived Conics Christopher J Bradley
A
F2
E3
E U F E1
P W V F1
B
D3
D
D2
C
Fig. 1 A Conic derived from Cevians 1. Introduction In Fig. 1 the Cevians AD, BE, CF of triangle ABC are drawn through a point P. Lines through E, F parallel to AD meet BC at points D2 and D3 respectively. Points E3, E1 on CA and F1, F2 on AB are defined similarly. The Cevian Conic is known to pass through these six points. We show that 1
this conic is a circle if and only if P coincides with the orthocentre H of ABC. It is also shown that in all cases F2E3, D3F1, E1D2 are parallel to the sides of ABC and the lines E1F1, F2D2, D3E3 form the medial triangle UVW of triangle DEF. Working is carried out using areal co-ordinates with ABC as triangle of reference. 2. The co-ordinates of the six points We take P to have co-ordinates (l, m, n). It follows that D, E, F have co-ordinates D(0, m, n), E(l, 0, m), F(l, m, 0). Lines parallel to AD have the form yn – zm = k(x + y + z). The value of k if this line passes through F is k = mn/(l + m), so that its equation is mnx – nly + m(l + m + n)z = 0. (2.1) This line meets BC at the point D3 with co-ordinates (0, m(l + m + n), nl). It follows by cyclic change of x, y, z and l, m, n that the co-ordinates of E1 are (lm, 0, n(l + m + n)) and F2 are (l(l + m + n), mn, 0). The line parallel to AD through E has k = – mn/(n + l) and so has equation mnx + n(l + m + n)y – lmz = 0.
(2.2)
This line meets BC at the point D2 with co-ordinates (0, lm, n(l + m + n)). It follows by cyclic change of x, y, z and l, m, n that the co-ordinates of E3 are (l(l + m + n), 0, mn) and F1 are (nl, m(l + m + n), 0). 3. Results (i) (ii)
Inspection of the co-ordinates of F2 and E3 shows that the displacement F2E3 is (0, – mn, mn) which is therefore parallel to BC.# EF has equation mnx – nly – lmz = 0. (3.1) E3D3 has equation m2nx + l2ny – lm(l + m + n)z = 0. (3.2)
These two lines meet at the point U with co-ordinates (l(2l + m + n), m(n + l), n(l + m)), which is the midpoint of EF. (iii) The conic D3D2E1E3F2F1 has equation ux2 + vy2 + wz2 +2fyz + 2gzx + 2hxy = 0, where u = 2m2n2(l + m + n), 2
(3.3)
v = 2n2l2(l + m + n), w = 2l2m2(l + m + n), f = – lmn(2l2 + m2 + n2 + 2(lm + mn + nl)), g = – lmn(l2 + 2m2 + n2 + 2(lm + mn + nl)), h = – lmn(l2 + m2 + 2n2 + 2(lm + mn + nl)).
(3.4)
This conic is a circle if and only if (v + w – 2f)/a2 = (w + u – 2g)/b2 = (u + v – 2h)/c2. If we therefore set a2 = l(m + n), b2 = m(n + l), c2 = n(l + m) these equations and equations (3.4) are satisfied and conversely. This means 2lm = a2 + b2 – c2 which is proportional to 1/n. The conic is therefore a circle if and only if (l, m, n) = (1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)), that is if and only if P is the orthocentre H.
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Article: CJB/82/2010
Conics generated by Points on a Curve of degree Five Christopher J Bradley F3
A
F2 E3 E F X P
F1
B
D3
E1
D1
D2 C
D
To E2
Fig. 1 Perpendiculars on to the sides from the feet of Cevians 1
1. Introduction In Fig. 1 above P is a point not on the sides or their extensions of a triangle ABC. With P as Cevian point the feet of the Cevians AP, BP, CP are denoted by D, E, F respectively. From D perpendiculars DF1 and DE1 are drawn on to the sides AB, AC respectively. Points F2, D2 are similarly obtained from perpendiculars from E and points D3, E3 are obtained from perpendiculars from F. The condituion for the points D2, E1, E3, F2, F1, D3 to lie on a conic are investigated and it turns out that a conic is obtained if and only if P lies on a curve of degree five. Well known points lying on this curve are O, H, I, the circumcentre, orthocentre and incentre of ABC, and also the excentres. Unfortunately I do not have any software to draw such a curve. If F2E3 meets BC at D1 and E2, F3 are similarly defined it is obvious that D1E2F3 is a straight line. 2. Co-ordinates of main points We take P to have areal co-ordinates (l, m, n) so that the normalized co-ordinates of D, E, F are as follows: D(0, m/(m + n), n/(m + n)), E(l/(n + l), 0, n/(n + l), F(l/(l + m), m/(l + m)). It is now necessary to find the feet of the perpendiculars from D, E, F on to the other sides. The formula to be used is given in Bradley [1] and it yields the following results. F1(r, 1 – r, 0), where r = n(c2 + a2 – b2)/(2c2(m + n)), E1(1 – q, 0, q), where q = (c2m – a2m + b2(m + 2n))/(2b2(m + n)), D2(0, p, (1 – p)), where p = l(a2 + b2 – c2)/(2a2(l + m)), F2(r. 1 – r, 0), where r = ((a2 – b2)n + c2(2l + n))/(2c2(n + l)), D3(0, p, (1 – p)), where p = (a2(l + 2m) + l(b2 – c2))/(2a2(l + 2m)), E3(1 – q, 0, q), where q = m(b2 + c2 – a2)/(2b2(l + m)).
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
3. Condition for these six points to lie on a conic This turns out to be a very lengthy expression of degree five in l, m, n and of degree eight in a, b, c. This means that there is a curve of degree five in the plane of the triangle ABC such that if P(l, m, n) lies on the curve then the six points are co-conic. It has been checked that the circumcentre O, the orthocentre H, the incentre I and the three excentres lie on the curve (and the centroid, nine-point-centre, the symmedian point and Gergonne’s point do not). The equation of the curve is: (a8m2n(l – n)(m – n) + a6m(c2m(l3 + l2(3m – n) + nl(3n – 2m) – n2(m + 3n)) – b2n(l3 + 2l2n + l(3m2 + 3mn – 5n2) – mn(m + 3n))) + a4(b4n(l3(3m + 2n) + 2l2n(2m + 3n) + lm(3m2 + 9mn + 5n2) + m2n(m – n)) + b2c2m(l3(m – 3n) – l2(5m2 + 9mn – 6n2) + nl(m2 – 6mn – 9n2) – 2mn2(3m + n)) – c4m2(3l3 + l2(5m – 3n) + 3ln2 – n2(5m + 3n))) – a2(b6n(l3(3m + 4n) + 2l2n(m + 2n) + lm(m2 + 5mn + 9n2) + m2n(m + 3n)) + b4c2(l3(5m2 + 6mn – 4n2) – l2(m3 + 5m2n – 4n3) + lmn2(5m – 6n) + m2n2(m – 5n)) – b2c4m(l3(2m + 9n) + 6l2(m2 + mn – n2) – ln(m2 – 9mn – 3n2) + mn2(5m – n)) – c6m2(3l3 + l2(m – 3n) + nl(2m + n) – n2(3m + n))) +l(b2 – c2)(b6n(l2(m + 2n) – 2ln2 – mn2) + b4c2(l2(3m2 + 10mn + 6n2) + l(m3 + 5m2n + 2
2mn2 + 2n3) + mn(m2 + 2n2)) + b2c4m(l2(4m + 5n) – 2nl(2m + n) – n(2m2 + n2)) + c6m2(l2 – l(m + n) + mn))) = 0. (3.1)
Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath (2007).
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Article: CJB/2010/84
The G Circles and the Conic they determine Christopher J Bradley
V
A
N
P Q
G
W
L
C
B
U
M
Fig. 1 Circles through G and the Conic they determine
1. Introduction 1
In Fig.1 observe triangle ABC and three circles BGC.CGA, AGB through the centroid G. Each of these circles cuts the triangle in two additional points. Circle BGC meets CA at M and AB at W. Circle CGA meets BC at U and AB at N and circle AGB meets CA at V and BC at L. It is found that points U, V, W, L, M, N lie on a conic. We prove this and determine its equation using areal co-ordinates with ABC as triangle of reference. 2. The circles BGC, CGA, AGB and points L, M, N, U, V, W Since the co-ordinates of G are (1, 1, 1) it is actually very easy to determine the equations of the circles. Circle BGC has equation (a2 + b2 + c2)x2 – 3a2yz + (c2 + a2 – 2b2)zx + (a2 + b2 – 2c2)x = 0. (2.1) The equations of circles CGA, AGB maybe written down by repeated cyclic change of x, y, z and a, b, c. The co-ordinates of the six points are as follows: U: (0, 2a2 – b2 – c2, a2 + b2 + c2), V: (a2 + b2 + c2, 0, 2b2 – c2 – a2), W: (2c2 – a2 – b2, a2 + b2 + c2, 0), L: (0, a2 + b2 + c2, 2a2 – b2 – c2), M: (2b2 – c2 – a2, 0, a2 + b2 + c2), N: (a2 + b2 + c2, 2c2 – a2 – b2, 0).
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7)
Note that pairs L, U and M, V and N, W are isotomic conjugates 3. The conic UVWLMN The standard method of finding the equation of the conic was used, namely to start with the general equation of a conic and determine the coefficients that satisfy five of the six points and finally to check that the sixth point does actually lie on the conic. The equation turns out to be more complicated than one would imagine it to be and is (a2 + b2 + c2)(2a2 – b2 – c2)(2b2 – c2 – a2)(2c2 – a2 – b2)(x2 + y2 + z2) – (a2 + b2 – 2c )(c + a2 – 2b2)(5a4 – 2a2(b2 + c2) + (b2 + c2)2)yz – (b2 + c2 – 2a2)(a2 + b2 – 2c2) (5b4 – 2b2(c2 + a2) + (c2 + a2)2)zx – (b2 + c2 – 2a2)(a2 + b2 – 2c2)(5c4 – 2c2(a2 + b2) + (a2 + b2)2)xy = 0. (3.1) 2
2
4. Further developments It may be asked if any points other than G determine a conic by means of a similar construction. The answer appears to be that if P lies on a curve in the plane of ABC then a conic will be determined. At present I know by drawing with Cabri II plus that K the symmedian point determines not only a conic but in fact a Tucker circle with centre lying at a point Q on OK where OQ = 3/2 OK. Other points producing a conic are the two Fermat points and the two isodynamic points. It appears that the curve may be invariant under isogonal conjugation. The equation of the Tucker circle mentioned above is 2
ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(4.1)
u = 6b2c2(b2 + c2 – 2a2), v = 6c2a2(c2 + a2 – 2b2), w = 6a2b2(a2 + b2 – 2c2), f = – a2(a4 – a2(b2 + c2) – 2(b4 – 7b2c2 + c4)), g = – b2(b4 – b2(c2 + a2) – 2(c4 – 7c2a2 + a4)), h = – c2(c4 – c2(a2 + b2) – 2(a4 – 7a2b2 + b4)).
(4.2) (4.3) (4.4) (4.5) (4.6) (4.7)
where
Flat 4, Terrill Court, 12- 14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/85
A Converse of the Pascal Line Property Christopher J Bradley Abstract: ‘If pairs of points L,U on BC, M, V on CA and N, W on AB lie on a conic and if LU^WM = D, MV^UN = E, NW ^VL = F then D, E, F are collinear’ is a statement of the Pascal line property. In this brief article we provide a converse.
V
A
F E
W
M
N U
B
L
C
D
Fig.1 A Pascal line of a Conic 1. Introduction A Converse of the Pascal line theorem stated above is as follows: Let DEF be a transversal of a triangle ABC, with D on BC, E on CA and F on AB, and let points L, M, N lie on BC, CA, AB 1
respectively, and if we define U = NE^BC, V = LF^CA and W = MD^AB, then L, M, N, U, V, W lie on a conic. A pure geometrical proof is straightforward, but we give a proof using areal co-ordinates with ABC as triangle of reference in order to appreciate the equation of the conic in terms of the given transversal and the three specified points. 2. Given DEF and points L, M, N to find points U, V, W Suppose the transversal DEF has equation lx + my + nz = 0. Then D has co-ordinates (0, n, – m), E has co-ordinates (– n, 0, l) and F has co-ordinates (m, – l, 0). Now let the co-ordinates of L, M, N be L(0, p, 1 – p). M(1 – q, 0, q), N(r, 1 – r, 0). The equation of DM is nqx + (q – 1)(my + nz) = 0.
(2.1)
It follows that DM meets AB at the point W with co-ordinates W(m(1 – q), nq, 0). Similarly U has co-ordinates U(0, n(1 – r), lr) and V has co-ordinates (mp, 0, l(1 – p)). 3. The conic LMNUVW It is now standard work to find the equation of the conic through L, M, N, U, V and to check that W lies on it. The equation of the conic is q(r – 1)(p – 1)x2/m + r(p – 1)(q – 1)y2/n + p(q – 1)(r – 1)z2/l + (q – 1){(r – 1)(p – 1)/l + rp/n}yz + (r – 1){(p – 1)(q – 1)m + pq/l}zx + (p – 1){(q – 1)(r – 1)/n + qr/m}xy = 0. (3.1) The conditions for LMNUVW to be a circle are m(1 – q)(lr + n(1 – r)) = ka2, n(1 – r)(mp + l(1 – p)) = kb2, l(1 – p)(nq + m(1 – q)) = kc2, where a, b, c are the side lengths of ABC and k is some positive real constant.
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2
(3.2)
Article: CJB/2010/86
The Perpendiculars to Three Segments at a Point determine a Conic Christopher J Bradley F
A
V N W P
M B
L
U
C
D
E
Fig. 1 Perpendiculars to AP, BP, CP determine a Conic LMNUVW Abstract: In the plane of a triangle ABC a general point P is chosen. Lines perpendicular to AP and BP and CP meet the sides at W, M and U, N and V, L respectively. We prove that the points L, M, N, U, V, W always lie on a conic and that the third points of intersection D, E, F lie on a line.
1
1. Introduction The important result is that if the perpendiculars to AP, BP, CP through P meet sides BC, CA, AB respectively at D, E, F then DEF is a straight line. It then follows immediately from the result of Article : CJB/2010/85 [1], since DM passes through W on AB, EN passes through U on BC and FL passes through V on CA, that points L, M, N, U, V, W lie on a conic. We prove that DEF is a straight line, using areal co-ordinates with ABC as triangle of reference. 2. The displacements AP and WM Suppose P has co-ordinates (l, m, n), where l + m + n = 1. Then the displacement AP = (l – 1, m, n). Let the line WPM have equation ux + vy + wz = 0. We have to write down the conditions so that (i) P lies on the line and (ii) the displacement WM is perpendicular to AP. The first is easy enough and is ul + vm + wn = 0.
(2.1)
Suppose now M has co-ordinates (p, q, r) the p = – w/(u – w), q = 0 and r= u/(u – w). Similarly W has co-ordinates (– v/(u – v), u/(u – v), 0) and the displacement WM is (x, y, z), where x = u(v – w)/{(u – v)(u – w)}, y = u/(v – u), z = u/(u – w) (2.2) Writing f = l – 1, g = m, h = n condition (ii) that WM is perpendicular to AP is a2(gz + hy) + b2(hx + fz) + c2(fy + gx) = 0.
(2.3)
Values of (u, v, w) satisfying equations (2.1) and (2.3) are u = – 2a2mn – (l – m – n – 1)(b2n + c2m), v = a2n(l – m + n) + b2n(1 – 2l) + c2(l2 – l(m – n + 1) – n), w = a2m(l + m – n) + b2(l2 + l(m – n + 1) – m) + c2m(1 – 2l).
(2.4) (2.5) (2.6)
Unnormalised co-ordinates of M and N now follow and those of M are (x, y, z) where x = a2m(l + m – n) + b2(l2 + l(m – n – 1) – m) + c2m(1 – 2l). y = 0, z = 2a2mn + (l – m – n – 1)(b2n + c2m).
(2.7) (2.8) (2.9)
And those of W are (x, y, z), where x = a2n(l – m + n) + b2n(1 – 2l) + c2(l2 – l(m – n + 1) – n), y = 2a2mn + (l – m – n – 1)(b2n + c2m), z = 0.
(2.10) (2.11) (2.12)
3. The co-ordinates of D, E, F 2
The equation of the line WM may now be written down and it is found that it meets BC at the point D with co-ordinates (x, y, z), where x = 0, (3.1) 2 2 2 2 y = – (a m(l + m – n) + b (l + l(m – n – 1) – m) + c m(1 – 2l)), (3.2) 2 2 2 2 z = a n(l – m + n) + b n(1 – 2l) + c (l – l(m – n + 1) – n). (3.3) The co-ordinates of E, F follow by cyclic change of x, y, z and a, b, c and l, m, n. 4. DEF is a straight line The ratio BD/DC follows from equations (3.2) and (3.3) and similarly CE/EA and AF/FB may be obtained. It then follows, since l + m + n = 1, that (BD/DC)(CE/EA)(AF/FB) = – 1 and hence by Menelaus’ theorem that DEF is a straight line. That a conic passes through L, M, N, U, V, W follows at once from the result of Article 85.
Reference 1. C. J. Bradley, Article: CJB/2010/85 of this series.
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Article: CJB/2010/87
Two connected Rectangular Hyperbolae Christopher J Bradley
A S D O
G R
P E
C Q B
F
Fig. 1 The Rectangular Hyperbola through the Midpoints, Centre and Diagonal Points of a Cyclic Quadrilateral 1
1. Introduction Fig. 1 shows a Cyclic Quadrilateral and the Rectangular Hyperbola Σ through the midpoints P, Q, R, S of its sides and its centre O. It is shown that this conic Σ automatically passes through the diagonal points. This conic cuts the circle at four points defining another cyclic quadrilateral and its corresonding rectangular hyperbola is also investigated. Cabri II plus indicates that the centre of the circle and the centres of the two rectangular hyperbolae are collinear. We use Cartesian co-ordinates, origin O. 2. Co-ordinates of points P, Q, R, S and the equation of Σ We take the cyclic quadrilateral to have equation x2 + y2 = 1, with origin at its centre O. We use parameters to designate points so that A has parameter a and has co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)). The midpoint P of AB therefore has co-ordinates (x, y), where x = {(1 + ab)(1 – ab)}/{(1 + a2)(1 + b2)}, y = {(a + b)(1 + ab)}/{(1 + a2)(1 + b2)}.
(2.1) (2.2)
oints Q, R, S have similar co-ordinates but with parameters b, c and c, d and d, a respectively. The equation of the conic Σ through P, Q, R, S, O may now be found and is ux2 + vy2 + 2hxy + 2gx + 2fy = 0,
(2.3)
where u = 2(abc + abd + acd + bcd – a – b – c – d), v = – 2(abc + abd + acd + bcd – a – b – c – d), f = 2(abcd – 1), g = (abc + abd + acd + bcd + a + b + c + d), h = 2(abcd – ab – ac – ad – bc – bd – cd +1).
(2.4) (2.5) (2.6) (2.7) (2.8)
The conic Σ is known to be a rectangular hyperbola. 3. Σ passes through the diagonal points E, F, G It is suficient to show that Σ passes through E. The equation of AC is (1 – ac)x + (a + c)y = (1 + ac)
(3.1)
The equation of BD is similar with parameters b and d instead of a and c. The intersection of these lines is the diagonal point E with co-ordinates x = – (abc + acd – abd – bcd – a + b – c + d)/(abc + acd – abd – bcd + a – b + c – d), y = 2(ac – bd)/( abc + acd – abd – bcd + a – b + c – d). (3.2) It may now be checked that E lies on Σ, as do F and G by symmetry of parameters. 2
4. The second rectangular hyperbola
K
A
S D G P
O N
R E
L
C
Q M B
F
Fig. 2 The two Rectangular Hyperbolae We suppose now that Σ meets the circle again at points K, L, M, N with parameters k, l, m, n. Subsituting x = (1 – t2)/(1 + t2), y = 2t/(1 + t2) into equation (2.3) we find that k, l, m, n are the roots of the quartic equation (a + b + c + d)t4 – 2t3(ab + ac + bc + ad + bd + cd – 2) + 3t2(abc + abd + acd + bcd – a – b – c – d) – 2t(2abcd – ab – ac – ad – bc – bd – cd) – (abc + abd + acd + bcd) = 0. (4.1)
3
From equation (4.1) it is possible to deternie the symmetric functions of k, l, m, n. From here it is a matter of calculation using DERIVE to work out the equation of the rectangular Hyperbola passing through O and the midpoints of the sides of the cyclic quadrilateral KLMN. The result is 2(abcd – ab – ac – ad – bc – bd – cd + 1)x2 – 2(abcd – ab – ac – ad – bc – bd – cd + 1)y2 – 4(abc + abd + acd + bcd – a – b – c – d)xy – 2(abcd – 1)x – (abc + abd + acd + bcd + a + b + c + d)y = 0. (4.2)
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Article: CJB/88/2010
A Singular Cyclic Quadrilateral Christopher J Bradley
L
X
To L To X
A Z M B
E
H D T K
C N
V
Y
R
P G
To F
Q
To U To S
S Fig. 1 A Cyclic Quadrilateral with a Special Property
Abstract: A Cyclic Quadrilateral ABCD is such that the tangents at A and C and the line BD are concurrent. It follows that the tangents at B and D and the line AC are also concurrent and many other lines not normally present are defined.
1
F
1. Setting the Scene The number of lines and points in the configuration can increase endlessly, so we restrict our considerations as we now describe. The cyclic quadrilateral ABCD is such that AC and BD meet at E, AB and CD meet at F and AD and BC meet at G defining the diagonal point triangle EFG as usual. The tangents at A and C meet at P and the tangents at B and D meet at Q. It is known that P and Q lie on FG. The special property possessed by ABCD is that its diagonal AC passes through Q. We prove that in this case the diagonal BD passes through P. The following points are also defined. The tangents at A and B meet at L, the tangents at A and D meet at M, the tangents at B and C meet at N, the tangents at C and D meet at K. AB and the tangent at C meet at Y, AB and the tangent at D meet at X, BC and the tangent at A meet at U, BC and the tangent at D meet at V, CD and the tangent at A meet at H, CD and the tangent at B meet at R, DA and the tangent at B meet at S and the line DA meets the tangent at C at T. Finally to provide some variety, the point Z is defined as the intersection of BD and FL. Of the eleven dotted lines in Fig. 1 we show that seven would appear for any cyclic quadrilateral, but that four appear in this case when ABCD is special. 2. Co-ordinates and the condition for ABCD to be special We use Cartesian co-ordinates with circle ABCD the unit circle. Points on the circle A, B, C, D have parameters a, b, c, d and co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)) etc. The special condition we impose provides a relation that is linear in each of the variables a, b, c, d. We do not keep on writing down the equations of chords and tangents of ABCD, but simply remark that the equation of the line AB is (1 – ab)x + (a + b)y = (1 + ab), (2.1) and that the tangent at A has the equation (1 – a2) + 2ay = (1 + a2).
(2.2)
Chords AB and CD meet at F with co-ordinates (x, y), where x = – (abc + abd – acd – bcd – a – b + c + d)/(abc + abd – acd – bcd + a + b – c – d), y = 2(ab – cd)/ (abc + abd – acd – bcd + a + b – c – d).
(2.3) (2.4)
By exchanging b and d in equations (2.3) and (2.4) one gets the co-ordinates of G and by exchanging b and C in equations (2.3) and (2.4) one obtains the co-ordinates of E. The tangents at A and C meet at the point P with co-ordinates (x, y), where x = (1 – ac)/(1 + ac), y = (a + c)/(1 + ac).
2
(2.5) (2.6)
By using b, d rather than a, c in equations (2.5) and (2.6) one gets the co-ordinates of Q. It may now be shown that the lines FG and PQ have identical equations, namely (abc – abd + acd – bcd – a + b – c + d)x + 2(bd – ac)y + (abc – abd + acd – bcd + a – b + c – d) = 0.
(2.7)
The condition that ABCD is special, namely that A, C, Q are collinear is a = {– c(b + d) + 2bd}/(b + d – 2c).
(2.8)
The symmetric nature of this equation implies that B, D, P are also collinear. 3. The co-ordinates of 14 points The 14 points concerned are defined in section 1 and may be identified in Fig. 1. The chord AB meets the tangent at C at the point Y with co-ordinates (x, y), where x = – (2abc – (a + b)c2 – a – b + 2c)/(2abc – (a + b)c2 + a + b – 2c), y = 2(ab – c2)/(2abc – (a + b)c2 + a + b – 2c).
(3.1) (3.2)
The co-ordinates of X, U, V, H, R, S, T may now be written down from equations (3.1) and (3.2) by means of the following: For X use a, b, d instead of a, b, c. For U use b, c, a instead of a, b, c. For V use b, c, d instead of a, b, c. For H use c, d, a instead of a, b, c. For R use c, d, b instead of a, b, c. For S use d, a, b instead of a, b, c. For T use d, a, c instead of a, b, c. The co-ordinates of the point P are given by equations (2.5) and (2.6) and the co-ordinates of L, M, N, K may be written down from them by means of the following: For L use a, b instead of a, c. For M use a, d instead of a, c. For N use b, c instead of a, c. For K use c, d instead of a, c. 4. Results The method used is as follows: to prove that four or more points are collinear we find the equation of the line joining two of them and check the remaining points lie on that line and to prove that three points are collinear we use the determinant of their co-ordinates and show it to be zero. Results are as follows: (i) L, E, K, G are collinear on the line with equation (abc + abd – acd – bcd – a – b + c + d)x + 2(cd – ab)y + (abc + abd _ acd – bcd + a + b – c – d) = 0; (4.1) (ii) E, T, U are collinear; 3
(iii) (iv) (v) (vi) (vii) (viii) (ix)
E, V, S are collinear; G, H, X are collinear; Q, T, H are collinear; P, V, R are collinear; G, R, Y are collinear; T, V, F are collinear; M, E, N, F are collinear on the line with equation (abc – abd – acd + bcd + a – b – c + d)x + 2(ad – bc)y + (abc – abd – acd + bcd – a + b + c – d) = 0; (4.2) Of these collinearities all would appear with a general cyclic quadrilateral, except (v) and (vi) which only occur when ABCD is special. 5. The point Z and two more collinearities The point Z is the intersection of BD and FL and is introduced because it adds some variety to what has preceded section 5 and because it produces collinearities that only occur when ABCD is special. They are (x) Q, Y, Z; (xi) G, N, Z.
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/89
Perpendiculars in a Cyclic Quadrilateral Christopher J Bradley
A Q W Y N X
D
O
To K
S G U
R
C
M
P B H
L To F
J
Fig. 1 Perpendiculars drawn in ABCD produce four more Cyclic Quadrilaterals Abstract: Perpendiculars from A and C on to opposite sides of a cyclic quadrilateral produce four more cyclic quadrilaterals and two sets of parallel lines one set containing five lines and the other set containing six lines. 1. Introduction 1
To be clear about the configuration in Fig.1 it needs to be described in the order in which the various lines and circles are drawn. First drawn is a general cyclic quadrilateral ABCD. Then perpendiculars from A are drawn onto BC and CD meeting them at P and Q respectively. Similarly perpendiculars from C on to AB and DA are drawn meeting them at R and S respectively. It is the found that PQRS is a circle centre X passing through A and C. Point U is the intersection of AP and CR and point W is the intersection of AQ and CS. It is now found that AUCW is cyclic with circle centre Y on OX produced, where OX = XY. Also BPUR and DQWS are cyclic quadrilaterals whose circles are of equal radius. Also cyclic quadrilaterals AUCW and ABCD have equal angles (with angle B = angle W etc.). External diagonal points are now found. AB and DC meet at F and AD and BC meet at G. AU and WC meet at H and AW and UC meet at K. Further points in the figure are (i) J is the intersection of AB and WC, (ii) M is the intersection of AD and UC, (iii) L is the intersection of AU and DC, (iv) N is the intersection of AW and BC. The construction ensures that the following are straight lines: ARBJ, AUPHL, ADSGM, AQWK, RUCMK, and BPCG. Remarkably from the points defined the following lines are parallel: BD, JL, MN, PS, RQ and UW and the following lines also parallel: OXY, BU, DW, HG and FK. It also transpires that JL = MN. In the following sections we prove these results using Cartesian co-ordinates with circle ABCD the unit circle, though the right angles are used in dealing with three of the cyclic quadrilaterals. 2. ABCD and the points P, Q, R, S and the circle PQRS We take A to have co-ordinates (0, 1) and B, C, D to have parameters b, c, d so that B, for example, has co-ordinates ((1 – b2)/(1 + b2), 2b/(1 + b2)). The equation of the line AB is (1 – b)x + (1 + b)y = 1 + b. (2.1) The equation of the line perpendicular to AB through C therefore has equation (1 + c2)((1 + b)x – (1 – b)y) = (1 – c2)(1 + b) – 2c(1 – b).
(2.2)
The point R where these two lines meet has co-ordinates (x, y), where x = {(1 – c)(b2c + b(1 + c) + 1)}/{(1 + b2)(1 + c2)}, y = {b2(1 + c) + b(1 – c)2 + c(1 + c)}/{(1 + b2)(1 + c2)}.
(2.3)
Similarly the point S has co-ordinates (x, y), where x = – {(1 + d)(c2d + c(1 – d) – 1)}/{(1 + c2)(1 + d2)}, y = {c2(1 + d) + c(1 – d)2 + d(1 + d)}/{(1 + c2)(1 + d2)}.
(2.4)
And the point P has co-ordinates (x, y), where 2
x = {(1 – c)(b2c – b(1 + c) + 1)}/{(1 + b2)(1 + c2)}, y = {b2c(1 + c) + b(1 – c)2 + 1 + c)}/{(1 + b2)(1 + c2)}.
(2.5)
And finally the point Q has co-ordinates (x, y), where x = {(1 – d)(c2d – c(1 + d) + 1)}/{(1 + c2)(1 + d2)}, y = {c2d(1 + d) + c(1 – d)2 + d + 1}/{(1 + c2)(1 + d2)}.
(2.6)
It is, of course, immediate from the diagram, because of the right angles that the midpoint X of AC is the centre of the circle PQRS passing through A and C. This also follows by calculation and the equation of circle PQRS is (x2 + y2)(1 + c2) – (1 – c2)x – (1 + c)2y + 2c = 0. (2.7) The co-ordinates of X, the centre of circle PQRS are X((1 – c2)/{2(1 + c2)}, (1 + c)2/{2(1 + c2)}. (2.8) The radius of the circle PQRS is |1 – c |/√(2(1 + c2)). 3. Circle AUCW Simple angle chasing shows that AUCW is a cyclic quadrilateral, and also that its angles are the same as those of ABCD (with angle B = angle W and angle D = angle U). However we wish to obtain its equation. U is the intersection of AP and RC whose equations are respectively (b + c)x + (bc – 1)(y – 1) = 0, (3.1) and (b + 1)(1 + c2)x + (b – 1)(1 + c2) + b(c2 – 2c – 1) + c2 + 2cv – 1 = 0. (3.2) So U has co-ordinates (x, y), where x = 2(1 – b2c2)/{(1 + b2)(1 + c2)}, y = {(1 + b2)(1 + c)2 + 2b(1 + c2)}/{(1 + b2)(1 + c2)}.
(3.3) (3.4)
Similarly W has co-ordinates (x, y), where x = 2(1 – c2d2)/{(1 + c2)(1 + d2)}, y = {(1 + c2)(1 + d)2 + 2c(1 + d2)}/{(1 + c2)(1 + d2)}.
(3.5) (3.6)
The equation of circle AUCW may now be obtained and is (1 + c2)(x2 + y2) – 2(1 – c2)x – 2(1 + c)2y + c2 + 4c + 1 = 0.
(3.7)
3
The centre Y of circle AUCW is Y, with co-ordinates ((1 – c2)/(1 + c2), (1 + c)2/(1 + c2)). It follows that O, X, Y are collinear and OX = XY. The radius of circle AUCW is 1, the same as circle ABCD. However the two cyclic quadrilaterals are not similar. The gradient of OXY is (1 + c)/(1 – c) and both DW and BU have the same slope and are all therefore parallel. Simple angle chasing shows that BPUR and DQWS are equiangular and though BU = DW, these cyclic quadrilaterals are not congruent. 4. The points F, G, K, H, L, M, J, N These eight points are defined as follows F = AB^CD, G = AD^BC (the exterior diagonal points of ABCD), K = AW^UC, H = AU^CD (the exterior diagonal points of AUCW), L = AU^CD, M = AD^UC, J = AB^WC and N = BC^AW. We give the co-ordinates of these eight points: F: ((1 – d)(b + 1)(c – 1), 2(cd – b))/(bcd – bc – bd + cd – b + c + d – 1); (4.1) G: ((1 – c)(b – 1)(d + 1), 2(bc – d))/bcd + bc – bd – cd + b + c – d – 1): (4.2) 2 3 2 3 2 K: ((2(bc + 1)(c d – cd – c + 1), (b(c (d + 1) + c (d – 3) – c(d + 1) – d – 1) + c (d + 1) + c (d + 1) + c(3d – 1) – d – 1))/((1 + c2)(b(c(d – 1) – d – 1) + c(d + 1) + d – 1); (4.3) 2 3 2 3 H: ((2(1 – bc)(c d + c – cd – 1), (b(c (d + 1) + c ( d + 1) + c(3 – d) – d – 1) + c (d + 1) + c2(1 – 3d) – c(d + 1) – d – 1))/((1 + c2)(b(c(d + 1) – d + 1) + c(1 – d) – d – 1); (4.4) 2 2 2 L: ((1 – d)(c – 1)(bc – 1), b(c d + c(d – 1) + 1) + c d + c(1 – d) + 1)/(1 + c )(bd + 1); (4.5) 2 2 2 M: (((1 – c)(d + 1)(bc + 1), b( c + c(d – 1) + d) + c + c(1 – d) + d)/(1 + c )(bd + 1); (4.6) J: (– (1 + b)(c2d + c(1 – d) – 1), b(c(d – 1) + d + 1) + c(c(d + 1) – d + 1)/(1 + c2)(bd + 1);(4.7) N: (1 – b)(c2d – c(d + 1) + 1), bc(c(d + 1) + d – 1) + c(1 – d) + d + 1)/(1 + c2)(bd + 1); (4.8) 5. The two sets of parallel lines The gradients of the various lines may now be calculated and it is found that the following five lines OXY, BU, DW, HG, FK have gradients (1 + c)/(1 – c), so these lines are parallel to each other all being perpendicular to AC. These gradients are independent of the positions of B and D. It is also found that the following lines BD, JL, MN, PS, RQ, UW have gradients(bd – 1)/(b + d) and are therefore parallel to one another in the direction of BD. Interestingly MN and JL are not only parallel but are equal displacements. Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP 4
Article: CJB/2010/90
Perpendiculars from the vertices of a Cyclic Quadrilateral Christopher J Bradley
N
To N P
2 Z W
E
Y V
3 A H
Q O
4
U
R E'
X
M 1
C S Z'
B T FTo K
Fig. 1 1
G
D
K
The three Cyclic Quadrilaterals and their diagonal points Abstract: The perpendiculars from A to AB and DA and the six similar perpendiculars from B, C, D to the sides of a cyclic quadrilateral create two more cyclic quadrilaterals that are coaxal with ABCD. Their sides and their diagonal point lines exhibit some surprising properties. 1. Introduction The perpendiculars from A to AB and from B to BC meet at Q, the perpendiculars from B to BC and from C to CD meet at R, the perpendiculars from C to CD and from D to DA meet at S and the perpendiculars from D to DA and from A to AB meet at P. It is proved that PQRS is a cyclic quadrilateral. The perpendiculars from B to AB and from C to BC meet at U, the perpendiculars from C to BC and from D to CD meet at V, the perpendiculars from D to CD and from A to DA meet at W and the perpendiculars from A to DA and from B to AB meet at T. It is proved that TUVW is a cyclic quadrilateral. It is also proved that the three circles ABCD, PQRS, TUVW are coaxal. The external diagonal points of ABCD are F and G. Those of PQRS are H and K and those of TUVW are M and N. It is proved that KFN is a straight line passing through O, the centre of ABCD, and that HGM is a straight line also passing through O. It is also proved that HKMN is a parallelogram. Cartesian co-ordinates are used throughout with ABCD the unit circle. 2. The co-ordinates of the vertices of the three circles and the equations of the circles With ABCD as unit circle, we give A, B, C, D parameters a, b, c, d so that the co-ordinates of A are ((1 – a2)/(1 + a2), 2a/(1 + a2)) and similarly for B, C, D with appropriate change of parameter. The equation of the perpendicular to AB at A is (a + b)x – (1 – ab)y = b – a.
(2.1)
The equation of the perpendicular to DA at D is (d + a)x – (1 – da) = a – d.
(2.2)
These two lines meet at P whose co-ordinates are (x, y), where x = {(1 + a2)(b + d) – 2a(1 + bd)}/{(1 + a2)(b – d)}, y = 2(bd – a2)/{(1 + a2)(b – d)}.
(2.3) (2.4)
The co-ordinates of Q, R, S may now be written down from equations (2.3) and (2.4) by cyclic change of a, b, c, d. 2
The equation of circle PQRS may now be determined and is (a – c)(b – d)(x2 + y2) + 4(bd – ac)x + 2(abc + acd – abd – bcd – a + b – c + d)y – (a – c)(b – d) = 0.
(2.5)
The perpendicular to DA at A has equation (a + d)x – (1 – ad)y = d – a.
(2.6)
The perpendicular to AB at B is (b + a)x – (1 – ba)y = a – b.
(2.7)
These lines met at the point T with co-ordinates (x, y), where x = (1 + a2)(b + d) – 2a(1 + bd)/{(1 + a2)(d – b)}, y = 2(bd – a2)/{(1 + a2)(d – b)}.
(2.8) (2.9)
The co-ordinates of U, V, W may now be written down from equations (2.8) and (2.9) by cyclic change of a, b, c, d. The equation of circle TUVW may now be determined and is (a – c)(b – d)(x2 + y2) – 4(bd – ac)x – 2(abc + acd – abd – bcd – a + b – c + d)y – (a – c)(b – d) = 0.
(2.10)
It is immediately clear that circles PQRS and TUVW have the same radius and have centres Y and X respectively such X, O, Y are collinear and XO = OY. 3. The diagonal points F, G, H, K, M, N AB and CD meet at F whose co-ordinates are (x, y), where x = – (abc – acd + abd – bcd – a – b + c + d)/(abc – acd + abd – bcd + a + b – c –d), (3.1) y = 2(ab – cd)/(abc – acd + abd – bcd + a + b – c – d). (3.2) AD and BC meet at G whose co-ordinates may be written down from equations (3.1) and (3.2) by exchanging b and d. PQ is the perpendicular to AB at A and QR is the perpendicular to BC at B. Their equations, see section 2, are respectively (a + b)x – (1 – ab)y = b – a and (b + c)x – (1– bc)y = c – b. Similarly RS has equation (c + d)x – (1 – cd)y = d – c and SP has equation (a + d)x – (1 – ad)y = a – d. The intersection of PQ and RS is the point H whose coordinates are (x, y), where x = – (abc – abd – acd + bcd + a – b – c + d)/(abc + abd – acd – bcd + a + b – c –d), (3.3) y = 2(bc – ad)/(abc + abd – acd – bcd + a + b – c – d). (3.4) The intersection of QR and PS is the point K whose co-ordinates are (x, y), where x = (abc + abd – acd – bcd – a – b + c + d)/(abc – abd – acd + bcd – a + b + c – d), (3.5) 3
y = 2(cd – ab)/(abc – abd – acd + bcd – a + b + c – d).
(3.6)
WT is the perpendicular to DA at A and has equation, see section 2, (a + d)x – (1 – ad)y = d – a. Similarly TU has equation (a + b)x – (1 – ab)y = a – b, UV has equation (b + c)x – (1 – bc)y = b – c and VW has equation (c + d)x – (1 – bc)y = b – c. M is the intersection of TU and VW and has co-ordinates (x, y), where x = (abc – abd – acd + bcd + a – b – c + d)/(abc + abd – acd – bcd + a + b – c – d), (3.7) y = 2(ad – bc)/(abc + abd – acd – bcd + a + b – c – d). (3.8) The intersection of UV and TW is N whose co-ordinates are (x, y), where x = – (abc + abd – acd – bcd – a – b + c + d)/(abc – abd – acd + bcd – a + b + c –d), (3.9) y = 2(ab – cd)/(abc – abd – acd + bcd – a + b + c –d). (3.10) 4. Results involving the diagonal points The equation of FK is 2(ab – cd)x + (abc + abd – acd – bcd – a – b + c + d)y = 0.
(4.1)
It may now be checked that O and N lie on FK. The equation of GH is 2(ad – bc)x – y(abc – abd – acd + bcd + a – b – c + d) = 0.
(4.2)
It may now be checked that O and M lie on GH. The slopes of HK and NM may be worked out and the are found to be equal (but are technically very complicated and not worth recording). Similarly the slopes of HN and KM are equal to (abc + abd + acd + bcd + a + b + c + d)/{2(1 – abcd)}. It follows that the figure HNMK is a parallelogram. 5. The three circles are coaxal The equations of the three circles are x2 + y2 = 1 and those of equations (2.5) and (2.10). It is evident that they have a common chord whose equation is 2(ac – bd)x + (abc – abd + acd – bcd – a + b – c + d)y = 0. (5.1) It follows that they are part of a coaxal system.
Flat 4, Terrill Court, 12-14, Apsley Road. BRISTOL BS8 2SP 4
5
Article: CJB/2010/91
6 Conics Christopher Bradley Abstract: When a circumconic is drawn to triangle ABC the tangents at A, B, C form a triangle DEF which is in perspective with ABC. If the lines AD, BE, CF meet at U and meet the circumconic at R, S, T then the tangents at R, S, T form a triangle LMN, It is shown that D, E, F, L, M, N lie on a conic, Then, by drawing tangents and chords four other conics may be constructed with centres all lying on a line through U.
1. Introduction Let ABC be a triangle and let. Σ1 be an arbitrary circumconic of ABC. The tangents at A, B, C to Σ1 form a triangle DEF. We prove that AD, BE, CF meet at a point U, so there is a Desargues’ axis of perspective 11 22 33. Let AD, BE, CF meet Σ1 again at R, S, T. Now triangles ABC and RST intersect at six points 31 21 23 13 12 32 as shown in Fig.1 and we prove that these six points lie on a conic Σ2. The tangents at R, S, T to Σ1 form a triangle LMN and we prove D, E, F, L, M, N lie on a conic Σ3. FM and EN meet at G and H, K are similarly defined and it is shown that G, H, K lie on the perspectrix 11 22 33. Many other chords are shown to pass through the six points 11, 22, 33, G, H, K. The point 1 is defined as EN^AS and the point 1' is defined as FM^BR. Points 2, 2', 3, 3' are similarly defined by cyclic change of letters and numbers. It is then proved that points 1, 2, 3, 1', 2', 3' lie on a conic Σ4 which is an inscribed conic to DEFLMN. Points such as B, T, 2, 3 and A, T, 3', 1' are shown to be collinear. Finally twelve points such as md = MD^AT3'1' and mf = MF^CR31 are shown to lie on a conic Σ5. The five conics are shown to have centres on the same line through U. Many other sets of collinear points are shown to exist. See Fig.1. Areal co-ordinates are used throughout with ABC as triangle of reference.
1
fm
lf To le
FL Z
L LE en 2
Q 3' X
F
A E
Y T
fl 1'
31 32
M mf
12 R
B 3 DM
13
D
dn
S
el
1 nd
C 2'
To 11
EN
23
U
MF md
21 W V
N
NDne dm
To 33 To G
To H
K 22
G 33
Fig.1 The six Conics . 2. Points D, E, F, the line of centres and points R, S, T We take the circumconic Σ1 to have equation fyz + gzx + hxy = 0. (2.1) The centre W of this conic has co-ordinates (f(g + h – f), g(h + f – g), h(f + g – h)). (If the conic is a circle then f = a2, g = b2, h = c2 and the co-ordinates of W will be recognized as those of the circumcentre.) 2
The equations of the tangents at A, B, C are gz + hy = 0, hx + fz = 0, fy + gx = 0.
(2.2) (2.3) (2.4)
The tangents at B, C meet at D (– f, g, h) and E, F have co-ordinates E(f, – g, h) and F(f, g, – h). It may now be checked that AD, BE, CF meet at a point U with co-ordinates U(f, g, h). The Desargues’ axis of perspective for triangles ABC and DEF is the line with equation x/f + y/g + z/h = 0. (2.5) The line UW has equation (g – h)x/f + (h – f)y/g + (f – g)z/h = 0.
(2.6)
This line will eventually be shown to contain the centres of the four other conics to be defined. The line AD has equation hy = gz and meets the circumconic at the point R with coordinates R (– f, 2g, 2h) Similarly S and T have co-ordinates S(2f, – g, 2h) and T(2f, 2g, – h). 3. The intersections of triangle ABC and RST and the conic Σ2 The equation of ST is ghx = 2hfy + 2fgz.
(3.1)
ST meets AB at the point 31 (3 for AB and 1 for the letter R not used whereas ST is used) with co-ordinate 31(2f, g, 0). Similarly ST meets AC at 21(2f, 0, h). Other points of intersection are 32(f, 2g, 0), 23(f, 0, 2h). 12(0, 2g, h), 13(0, g, 2h). The equation of the conic Σ2 containing these six points may now be calculated and is 2x2/f2 + 2y2/g2 + 2z2h2 – 5yz/(gh) – 5zx/(hf) – 5xy/(fg) = 0. (3.2) The centre V of this conic has co-ordinates (f(5g + 5h – 3f), g(5h + 5f – 3g), h(5f + 5g – 3h)
(3.3)
and V may be checked to lie on the line UW. 4. The points L, M, N and the conic Σ3 The tangents to the circumconic at R, S, T are as follows: 4ghx + hfy + fgz = 0, ghx + 4hfy + fgz = 0, ghx + hfy + 4fgz = 0.
3
(4.1) (4.2) (4.3)
The tangents at S and T meet at L with co-ordinates L(– 5f, g, h). Similarly M has coordinates (f, – 5g, h) and N has co-ordinates (f, g, – 5h). The equation of the conic Σ3 through D, E, F, L, M, N may now be calculated and is x2/f2 + y2/g2 + z2/h2 + 3yz/(gh) + 3zx/(hf) + 3xy/(fg) = 0. (4.4) The centre X of Σ3 is the point X with co-ordinates (f(3g + 3h – 4f), g(3h + 3f – 4g, h(3f + 3g – 4h)).
(4.5)
and X may be checked to lie on the line UW. 5. The tangents at D, E, F, L, M, N and the conic Σ4 We now record the equations of the tangents to Σ3 at the six points D, E, F, L, M, N. These are 2x/f + y/g + z/h = 0, (5.1) x/f + 2y/g + z/h = 0, (5.2) x/f + y/g + 2z/h = 0, (5.3) 2x/f + 5y/g + 5z/h = 0, (5.4) 5x/f + 2y/g + 5z/h = 0, (5.5) 5x/f + 5y/g + 2z/h = 0. (5.6) In what follows in this section we use the notation that, for example, FL stands for the intersection of the tangents at F and L to Σ3. There are six points of this type FL, LE, EN, ND, DM, MF and after recording their co-ordinates we show they lie on a conic Σ4. FL: (5f, g, – 3h), LE(5f, – 3g, h), EN(f, – 3g, 5h), ND(– 3f, g, 5h), DM(– 3f, 5g, h) and MF(f, 5g, – 3h). It may now be shown that these six points lie on a conic Σ4 with equation 13x2/f2 + 13y2/g2 + 13z2/h2 + 35yz/(gh) + 35zx/(hf) + 35xy/(fg) = 0.
(5.7)
The centre Q of this conic has co-ordinates (f(35g + 35h – 61f), g((35h + 35f – 61g),h((35f + 35g – 61h)).
(5.8)
It may now be checked that Q lies on the line UW. 6. The line GHK 11 22 33 We define the point G to be the intersection of FM and EN so G has co-ordinates (– 2f, g, h). Similarly H and K have co-ordinates H(f, – 2g, h) and K(f, g, – 2h). The equation of the line GHK is therefore x/f + y/g + z/h = 0.
(6.1)
Note that this is the Desargues’ axis of perspective for triangle ABC and DEF. 4
The equation of AD is hy = gz and G clearly lies on this line. Similarly H lies on BE and K on CF. The equation of BT is hx + 2fz = 0 and it is clear that G lies on BT. Similarly H lies on CR and K lies on AS. The equation of CS is gx + 2fy = 0 and G clearly lies on this line. Similarly H lies on AT and K lies on BR. The equation of 12 31 is ghx – 2hfy + 4fgz = 0.
(6.2)
Again G lies on this line and H lies on 23 12 and K lies on 31 23. The equation of 13 21 is ghx + 4hfy – 2fgz = 0.
(6.3)
Again G lies on this line and H lies on 21 32 and K lies on 32 13. We now investigate more closely the points 11, 22, 33, the points on the Desargues’ axis of perspective for triangle ABC and DEF. The equation of the tangent at C to the conic ABCRST is fy + gx = 0 and the tangent at T to conic ABCRST is ghx + hfy + 4fgz = 0 and these two lines pass through 33(– f,g, 0), which is also the intersection of AB and RS. 33, of course, lies on x/f + y/g + z/h = 0 as do 11(0, – g, h) and 22(f, 0, – h). 11 lies on the tangents at A and R to Σ1 and 22 lies on the tangents at B and S to Σ1. The tangent at F to Σ3, see equation (5.3), also passes through 33 and the tangents at D, E pass through 11, 22 respectively. Also the tangent at N to Σ3, see equation (5.6), passes through 33 and similarly the tangents at L, M to Σ3 pass through 11,22 respectively. 7. The points 1, 2, 3, 1', 2', 3' and the conic 1231'2'3' The point labelled 1 may be defined as the intersection of AS and CR. Their equations are 2hy + gz = 0 and 2gx + fy = 0. So 1 has co-ordinates 1(f, – 2g, 4h). The equation of NE is 2ghx + 3hfy + fgz = 0. (7.1) and it may be checked that 1 lies on this line. The equation of 32 23 is 2ghx – hfy – fgz = 0.
(7.2)
and it may be checked that 1 also lies on this line. Similarly the point 2(4f, g, – 2h) lies on BT, AS, LF and 13 31 and 3(– 2f, 4g, h) lies on CR, BT, MD and 21 12. The point labelled 1' may be defined as the intersection of AT and BR. Their equations are hy + 2gz = 0 and 2hx + fz = 0. So 1' has the co-ordinates (– f, – 4g, 2h). 1' also lies on FM, BT, AR and 32 23. Likewise 2'(2f, – g, – 4h) lies on BR, CS, DN, and 13 31. And 3'(– 4f, 2g, – h) lies on CS, AT, EL and 21 12. 5
It may now be shown that a conic Σ5 passes through 1, 2, 3, 1', 2', 3' and its equation is 2x2/f2 + 2y2/g2 + 2z2/h2 + 7yz/(gh) + 7zx/(hf) + 7xy/(fg) = 0. (7.3) The centre Y of Σ5 has co-ordinates (f(7g + 7h – 11f), g(7h + 7f – 11g), h(7f + tg – 11f)) It may be checked that Y lies on the line UW. The tangent at 1 to Σ5 passes through E and N, the tangent at 2 passes through F and L and the tangent at 3 passes through D and M. Similarly the tangent at 1' to Σ5 passes through F and M, the tangent at 2' passes through D and N and the tangent at 3' passes through E and L. It follows that Σ3 and Σ5 are the circumconic and inconic of a porism involving hexagons of which LFMDNE is a member. 8. The twelve point conic Σ6 The main difficulty in having twelve points to define is to provide a useful notation and we have chosen one in which the twelve points are divided into four groups of three, each group of three exhibiting a cyclic pattern so that one only needs to exhibit the lines passing through the leading point of each group and the points are also organized so that their coordinates may also be obtained by cyclic change of x, y, z and f, g, h. This has been made possible because of the care chosen with the labelling of points from the outset. The first three points are labelled le, mf, nd. The leading member le lies on LE and is on the extension of EL. It also lies on CA and it may be checked that it lies on 23BT and the tangents at F and 3' to the conics and Σ5 respectively. The equation of LE is ghx + 3hfy + 2fgz = 0, (8.1) and since le also lies on CA, the co-ordinates of le are (2f, 0, – h). The tangents at 3' and F are respectively x/f + 3y/g + 2z/h = 0, (8.2) and x/f + y/g + 2z/h = 0, (8.3) and le clearly lies on these lines. Similarly mf lies on MF, AB, 31CR, and the tangents at D and 1' and has co-ordinates (– f, 2g, 0). And again nd lies on ND, BC, 12AS and the tangents at E and 2' and has co-ordinates (0, – g, 2h). The second set of three points is el, fm, dn. The leading member el lies on EL, ST, 31CR, and the tangents at 3' and N. Its co-ordinates are (2f, – 4g, 5h). Similarly fm has coordinates (5f, 2g, – 4h) and dn has co-ordinates (– 4f, 5g, 2h). The third set of three points is lf, md, ne. The leading member lies on LF, AB, 2'3'CS and the tangents at E and 2. Its co-ordinates are (– 2f, g, 0). Similarly md has co-ordinates (0, – 2g, h) and ne has co-ordinates (f, 0, – 2h).
6
The fourth set of three points is fl, dm, en. The leading member lies on FL, ST, 1'2'BR and the tangents at M and 2. Its co-ordinates are (2f, 5g, – 4h). Similarly dm has co-ordinates (– 4f, 2g, 5h) and en has co-ordinates (5f, – 4g, 2h). These twelve points lie on the conic Σ6 with equation 2x2/f2 + 2y2/g2 + 2z2/h2 + 5yz/(gh) + 5zx/(hf) + 5xy/(fg) = 0.
(8.4)
The centre Z of Σ6 has co-ordinates (f(5g + 5h – 9f), g(5h + 5f – 9g), h(5f + 5g – 9h)),
(8.5)
and it may be checked that Z lies on the line UW.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/92
Three special Cyclic Quadrilaterals Christopher J Bradley Abstract: In any cyclic quadrilateral ABCD if AB and CD meet at F and AD and BC meet at G then FG is one side of the diagonal point triangle. It is then always the case that the tangents at B and D and the tangents at A and C meet at points U and V respectively lying on FG. The first special cyclic quadrilateral is when AC passes through U. If the tangents at A and B meet at P and Q, R, S are similarly defined then the second special cyclic quadrilateral is when AC is perpendicular to BD and then P, Q, R, S are concyclic. The third special cyclic quadrilateral is when AC is parallel to BD and then again P, Q, R, S are concyclic.
P A S
E D
B
V
R Q
C
G U F
Fig. 1 The first special cyclic quadrilateral 1. Introduction In all three cases we use Cartesian co-ordinates with circle ABCD having equation x2 + y2 = 1. (1.1) 1
and A having co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)) etc. And in each case we find the condition on a, b, c, d for the quadrilateral to be special. 2. The first special cyclic quadrilateral The line AB has equation (1 – ab)x + (a + b)y = 1 + ab.
(2.1)
and similarly for other sides and diagonals of ABCD. AB and CD meet at F with co-ordinates (x, y), where x = – (abc + acd – acd – bcd – a – b + c + d)/(abc + abd – acd – bcd + a + b – c – d), y = 2(ab – cd)/(abc + abd – acd – bcd + a + b – c – d). (2.2) G has co-ordinates similar to those in (2.2) but with b and d interchanged. The equation of the line FG may now be worked out and is (abc – abd + acd – bcd – a + b – c + d)x + 2(bd – ac)y + (abc – abd + acd – bcd + a – b + c – d) = 0. (2.3) The equation of the tangent at A is (1 – a2)x + 2ay = 1 + a2,
(2.4)
and similarly for tangents at other points with appropriate change of parameter. It may now be checked that the tangents at B and D meet at a point U with co-ordinates (x, y), where x = (1 – bd)/(1 + bd), y = (b + d)/(1 + bd). (2.5) and then it can be checked that U lies on FG. Similarly the tangents at A and C meet at a point V lying on FG. So far this is true for any cyclic quadrilateral. What now makes it special is that U also lies on AC. The condition for this turns out to be (a + c)(b + d) = 2(ac + bd). (2.6) The symmetry of equation (2.6) shows that when U lies on FG then V also lies on FG. See Fig. 1 for an illustration of this case. 3. The second special cyclic quadrilateral The point P is the intersection of the tangents at A and B and has co-ordinates (x, y), where x= (1 – ab)/(1 + ab), y = (a + b)/(1 + ab). (3.1) The points Q, R, S have similar sets of co-ordinates with (a, b) replaced by (b, c), (c, d), (d, a) respectively. The conditions for P, Q, R, S to lie on a circle are either 2
S A D P
E
R
O B
C
K
Q
Fig. 2 The second special cyclic quadrilateral (i) (ii)
abcd + ab – ac + ad + bc – bd + cd + 1 = 0, abc – abd + acd – bcd + a – b + c – d = 0.
(3.1) (3.2)
The second special cyclic quadrilateral is when condition (i) holds, which is when the diagonals AC and BD are at right angles. This is illustrated in Fig. 2. 4. The third special cyclic quadrilateral The third special cyclic quadrilateral is when condition (ii) holds, which is when the lines AC and BD are parallel. In this case ABCD is such that, for example, A lies between B and C. This is illustrated in Fig. 3.
3
D
R
O
S C
B A P
Q
Fig. 3 The third special cyclic quadrilateral
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
4
Article: CJB/2010/93
Circles formed by an Isosceles Trapezium Christopher J Bradley Abstract Given an isosceles trapezium ABCD with AD parallel to BC, if tangents to the cyclic quadrilateral ABCD are drawn to produce six points of intersection, then six more circles may be drawn, all passing through the centre O of ABCD. Five of these circles are obvious but the fact that the sixth circle passes through O is an interesting result and in this paper a proof is given using Cartesian co-ordinates.
U
D S O
A
R C P
B
V
Q
Fig. 1 The Seven Circles 1
1. Introduction In Fig.1 ABCD is an isosceles trapezium with AD parallel to BC. The centre of the circle ABCD is labelled O and P, Q, R, S are defined as the intersections of that tangents at A, B and A, C and B, D and C, D respectively. U, V are defined as the intersections of the tangents at A, D and B, C respectively. It is immediate from the existence of right angles between radii and tangents that the following circles exist: (i) DUAO, (ii) AQCO, (iii) DRBO, (iv) APBO, (v) DSCO. Fig. 1 indicates that the (obvious) circle PQRS also passes through O. In this brief article we prove this by using Cartesian co-ordinates with ABCD as unit circle. 2. Tangents and Points We take A to have co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)) and B, C, D to have similar co-ordinates but with parameters b, c, d respectively. Since AD is parallel to BC we have (1 – ad)/(a + d) = (1 – bc)/(b + c),
(2.1)
which, on simplification yields the condition abc + bcd – abd – acd – a + b + c – d = 0.
(2.2)
The equation of the tangent at A is (1 – a2)x + 2ay = (1 + a2),
(2.3)
The tangents at B, C, D have similar equations but with parameters b, c, d respectively. P is the intersection of the tangents at A and B and so has co-ordinates ((1 – ab)/(1 + ab), (a + b)/(1 + ab)).
(2.4)
Similarly Q, R, S have co-ordinates with pairs of parameters (a, c), (b, d), (c, d). We aim to show that circle PQRS passes through O. 3. Circle PQRS We suppose that circle PQRS has an equation of the form x2 + y2 + 2gx + 2fy + k = 0
(3.1)
and the circle passes through O if, and only if k = 0. We insert into Equation (3.1) the co-ordinates of P, Q, R and get three equations for g, f and k. The value of k obtained is 2
(1 + b2)(a3(bc – bd – cd – 1) \+ a2(bcd + b + c – d) + a(bc – bd – cd – 1) + bcd + b + c – d) (3.2) and by virtue of Equation (2.2) this vanishes.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP
3
Article: CJB/2010/94
Generalization of the Steiner Point Christopher Bradley Abstract: The outer Steiner ellipse passes through A, B, C and the images L, M, N of those vertices in a rotation of 180o about the centroid G. The Steiner point is the fourth point of intersection of the outer Steiner ellipse and the circumcircle of ABC. A generalization is obtained by replacing G by another point P internal to the triangle ABC. But more occurs as circles AMN, BMN, CNL intersect at a point U that lies on both the circumcircle and the ellipse ABCLMN and U is the generalized Steiner point.
A V N
M
P
C
B
U
L
Fig. 1 Generalization of the Steiner ellipse and the Steiner point U 1
1. Introduction Given a triangle ABC and an internal point P points L, M, N are defined on the extensions of AP, BP, CP so that P is the midpoint of AL, BM, CN respectively. It is shown that an ellipse passes through the six points A, B, C, L, M, N. Furthermore we show circles AMN, BNL, CLM have a common point U lying on both the ellipse and the circumcircle of ABC. The symmetry then ensures that circles LBC, MCA, NAB have a common point V lying on the ellipse and on the circumcircle of triangle LMN. Also UPV is a straight line and UP = PV. Areal co-ordinates are used with ABC as triangle of reference. 2. The co-ordinates of L, M, N and the conic ABCLMN We suppose that P has co-ordinates (l, m, n), where l + m + n = 1. Then the points L, M, N have co-ordinates L(2l – 1, 2m, 2n), M(2l, 2m – 1, 2n), N(2l, 2m, 2n – 1). The equation of the conic ABCLMN may now be obtained. Since it passes through A, B, C it must be of the form fyz + gzx + hxy = 0. (2.1) Putting in Equation (2.1) the co-ordinates of L, M, N we obtain the values of f, g, h and bearing in mind that l + m + n = 1 we obtain l(1 – 2l)yz + m(1 – 2m)zx + n(1 – 2n)xy = 0. ` (2.2) 3. The circles AMN, BNL, CLM The equation of any circle in areal co-ordinates is of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(3.1)
For a circle passing through A it is obvious that u = 0. Putting in the co-ordinates of M and N we obtain the appropriate values of v and w, which are v = 2(a2n(1 – 2n) + c2l(1 – 2m – 2n))/{(1 – 2m – 2n)(1 – 2l – 2m – 2n)}, w = 2(a2m(1 – 2m) + b2l(1 – 2m – 2n))/{(1 – 2m – 2n)(1 – 2l – 2m – 2n)}.
(3.2) (3.3)
Circles BNL, CLM may be obtained from Equations (3.2), (3.3) by cyclic change of u, v, w and l, m, n. Now the circumcircle has equation a2yz + b2zx + c2xy = 0. 2
(3.4)
The points of intersection of the conic and the circumcircle are A, B, C and the point U with coordinates (x, y, z) where x = 1/(b2n(2n – 1) + c2m(1 – 2m)), (3.5) 2 2 y = 1/(c l(2l – 1 ) + a n(1 – 2n)), (3.6) 2 2 z = 1/(a m(2m – 1) + b l(1 – 2l)), (3.7) where we have used l + m + n = 1 to secure the expected cyclic symmetry of these co-ordinates. Their form ensures that circles BNL and CLM also pass through U. It may also be checked that U lies on the ellipse ABCLMN. 4. Circles BCL, CAM, ABN and the point V The half turn rotational symmetry of the configuration about P now ensures that circles BCL, CAM, ABN pass through a common point V which lies on the circumcircle of LMN and the ellipse ABCLMN and that UPV is a straight line with UP = PV.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
3
Article: CJB/2010/95
A Nine Point Rectangular Hyperbola Christopher Bradley Abstract: The configuration consisting of the rectangular hyperbola passing through the incentre, the excentres, the centroid and deLongchamps point in a triangle and three other significant points exhibits some interesting properties which are investigated.
I3
A W
V
S
I2
M deL
N
G I E'
F' D'
R
B
To T
D
L
C Z
Y
P
F X
U
I1 E
Fig. 1 The nine point rectangular hyperbola 1. Introduction
1
Given a triangle ABC let the midpoints of the sides BC, CA, AB be denoted by L, M, N respectively. Points U, V, W on the extensions of AL, BM, CN respectively are defined so that AU = 2AL, BV = 2BM, CW = 2CN. The incentre I and the three excentres I1, I2, I3, the centroid G and deLongchamps point deL and the points U, V, W are the nine points we consider and first we prove that they lie on a conic, which is a rectangular hyperbola, since the incentre and the three excentres form an orthocentric quartet. The three lines UI1, VI2, WI3 form a triangle with vertices D, E, F which we show lie on BC, CA. AB respectively and it is proved that AD, BE, CF are concurrent at a point P. Points D', E', F' are the harmonic conjugates of D, E, F on BC, CA, AB respectively. It is a consequence that D', E', F' are collinear. It is also shown that lines AD, BE', CF' are concurrent at a point R, AD', BE, CF' are concurrent at a point S and AD', BE', CF are concurrent at a point T. If AD meets EF at X, then X is the harmonic conjugate of D' with respect to E and F and similarly for Y and Z. Areal co-ordinates are used throughout with ABC as triangle of reference. 2. The rectangular hyperbola through U,V,W, I, I1, I2, I3, G, deL The conic through the five points I, I1, I2, I3, G is easily obtained from their co-ordinates I(a, b, c), I1( – a, b, c), I2(a, – b, c), I3(a, b, – c), G(1, 1, 1) and is (b2 – c2)x2 + (c2 – a2)y2 + (a2 – b2)z2 = 0. (2.1) This conic clearly passes through U( – 1, 1, 1), V(1, – 1, 1), W(1, 1, – 1). Rather amazingly it passes through deLongchamps point deL whose x-co-ordinates is – 3a4 + 2a2(b2 + c2) + (b2 – c2)2 and whose y- and z- co-ordinates follow by cyclic change of a, b, c. The conic is a rectangular hyperbola, since the incentre and the three excentres form an orthocentric quartet. 3. The points D, E, F and their properties The point D is the intersection of lines VI2 and WI3. The equations of these lines are (b – c)x + (a – c)y + (a – b)z = 0,
(3.1)
and (b – c)x + (c – a)y + (b – a)z = 0.
(3.2)
The point D therefore has co-ordinates D(0, a – b, c – a). Similarly E has co-ordinates E(a – b, 0, b – c) and F has co-ordinates (c – a, b – c, 0). These points lie on the sides BC, CA, AB respectively and evidently AD, BE, CF are concurrent at P with co-ordinates P(1/(b – c), 1/(c – a), 1/(a – b)). (3.3)
2
4. The points D', E', F', R, S, T, X, Y, Z The point D' is the harmonic conjugate of D with respect to B and C and so has co-ordinates D'(0, b – a, c – a). Similarly E' and F' have co-ordinates E'(a – b, 0, c – b) and F'(a – c, b – c, 0). Since D, E, F are the feet of Cevians, the points D', E', F' are collinear. The point R is the point of concurrence of the three lines AD, BE', CF' and so has co-ordinates R(1/(c – b), 1/(c – a), 1/(a – b)). Similarly S and T have co-ordinates S(1/(b – c), 1/(a – c), 1/(a – b)), T(1/(b – c), 1/(c – a), 1/(b – a)). The point X is the intersection of AD with EF and since ABF and ACE are straight lines and the range {B, C; D, D'} is harmonic then it follows that D' and X separate E and F harmonically. Similarly for Y and Z.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
3
Article: CJB/2010/96
More on the Nine Point Rectangular Hyperbola Christopher Bradley
K W D
X
A
V W'
J
V'
T P G
O deL
E
S Y
U' B
A' F
C Z R
To C'
Q
Q' B'
I U
Fig. 1 Two triangles and a nine point rectangular hyperbola Abstract: The circumcircle of a triangle ABC is the Nine-Point circle of the triangle IJK of its excentres. The line JK passes through A and the midpoint D of JK, which lies therefore on the 1
circumcircle. With E, F similarly defined there are thus two triangles ABC and DEF and it is shown that their incentres and excentres lie on the nine point rectangular hyperbola discussed in CJB/2010/95. Other interesting properties emerge such as the deLongchamps point of DEF is the same point as the incentre of ABC. 1. Introduction A description of Fig. 1 and the order in which its elements were drawn is the best way of describing what happens and what we prove in the sections that follow. To start any general triangle ABC is drawn together with its incentre and excentres I, J, K. The incentre is labelled deL in Fig. 1 because (as we eventually show) it coincides with deLongchamps point of triangle DEF which we now define. The circumcircle of ABC is well known to be the nine-point circle of the triangle IJK of its excentres. It follows that the midpoints of the sides JK, KI, IJ lie on the circumcircle. These are the points D, E, F respectively. The incentre of DEF is the point P and its centroid is the point G. Its excentres are the points U, V, W. At this stage we are able to draw the rectangular hyperbola IJKdeLG, where deL is the incentre of triangle ABC and G the centroid of DEF. In the calculations that follow, which are carried out using areal co-ordinates with ABC as triangle of reference, it would be possible theoretically to find the equation of this conic. Unfortunately its equation is about half a page long and it would serve no purpose to write it down. The co-ordinates of D, E, F, however are easy enough to work out and to record. We now concentrate on triangle DEF. If U', V', W' are the midpoints of EF, FD, DE respectively then it turns out that I, J, K lie on DU', EV', FW' respectively and DI = 2DU', EJ = 2EV' and FK = 2FW'. Now, on referring back to CJB/2010/95, the conic IJKdeLG is seen to be the nine-point rectangular hyperbola of that article. It follows that the incentre P and excentres U, V, W lie on this rectangular hyperbola, these being the four additional points. So without obtaining the equation of the conic or indeed the co-ordinates of P, U, V, W we may identify these points for what they are. Additional properties of this configuration are as follows: (i) If BF^EC = R, CD^FA = S and AE^DB = T, then R, S, T are collinear, (ii) If BC^EF = A', CA^FD = B', AB^DE = C' then AA', BB', CC' are concurrent at a point Q lying on the circumcircle and (iii) DA', EB', FC' are concurrent at a point Q' lying on the circumcircle. 2. The points D, E, F and their properties Points I, J, K, the excentres of triangle ABC, have normalized co-ordinates I(– a, b, c)/(b + c – a), J(a, – b, c)/(c + a – b), K(a, b, – c)/(a + b – c). 2
(2.1)
D is the midpoint of JK and so has co-ordinates D(a2, bc – b2, bc – c2)/(a + b – c)(c + a – b).
(2.2)
E is the midpoint of KI and so has co-ordinates E(ca – a2, b2, ca – c2)/(b + c – a)(a + b – c).
(2.3)
F is the midpoint of IJ and so has co-ordinates F(ab – a2, ab – b2, c2)/(c + a – b)(b + c – a).
(2.4)
It may now be checked that D, E, F lie on the circumcircle with equation a2yz + b2zx + c2xy = 0.
(2.5)
The incentre of triangle ABC, deL (since we prove shortly that it is deLongchamps point of triangle DEF) has normalized co-ordinates (a, b, c)/(a + b + c). The co-ordinates of the centroid G of triangle DEF may now be worked out and are G(a(3a2 – b2 – c2 + 2bc – 2ca – 2ab), b(3b2 – c2 – a2 + 2ca – 2ab – 2bc), c(3c2 – a2 – b2 + 2ab – 2bc – 2ca)).
(2.6)
The conic Σ we now define as the conic IJKGdeL, but we do not record its equation as it is very long and complicated, and is not actually needed in this article. The circumcentre O of DEF (and indeed of ABC) has co-ordinates (a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2))/(2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4).
(2.7)
From O and G it is possible to work out the co-ordinates of deL and using the formula deL = 4O – 3G we find them to be (a, b, c)/(a + b + c) the same point as the incentre of triangle ABC. The midpoint U' of EF has normalised co-ordinates (a(2a2 – b2 – c2 + 2bc – ca – ab), b(2b2 – a2 – 2bc + ca – ab), c(2c2 – a2 – 2bc – ca +ab)) all divided by 2(a + b – c)(a – b – c)(c + a – b).
(2.8)
It may now be checked that the point on the extension of DU' twice as far from D as U' has coordinates (– a, b, c)/(b + c – a), which is I, the excentre of triangle ABC opposite A. From Article 95 (CJB/2010/95) it follows that the conic Σ also passes through P, U, V, W where P is the incentre and U, V, W are the excentres of triangle DEF, so that Σ is what we termed the nine point rectangular hyperbola. Again the co-ordinates of the points P, U, V, W are very long and complicated and are not recorded. 3
3. The Euler line of triangle DEF The cross over intersection of sides BF^CE = R has co-ordinates R(a(a – b)(a – c), b2(b – a), c2(c – a).
(3.1)
Similar points CD^AF = S has co-ordinates S(a2(a – b), b(b – c)(b – a), c2(c – b)) .
(3.2)
and AE^BD = t has co-ordinates T(a2(a – c), b2(b – c), c(c – a)(c – b).
(3.3)
These three points lie on the line with equation bc(b + c – a)(b – c)x + ca(c + a – b)(c – a)y + ab(a + b – c)(a – b)z.
(3.4)
It may now be checked that G and deL lie on this line, which is therefore the Euler line of triangle DEF. 4. The points A', B', C' and the points Q, Q' on the circumcircle The sides of ABC and DEF meet at the points BC^EF = A', CA^FD = B' and AB^ DE = C'. Their co-ordinates are A'(0, 1/(c(c – a)), 1/(b(a – b)), (3.5) B'(1/(c(b – c)), 0, 1/(a(a – b))), (3.6) C'(1/(b(b – c)), 1/(a(c – a)), 0). (3.7) The lines AA'. BB', CC' meet at the point Q with co-ordinates (a/(b – c), b/(c – a), c/(a – b)) and it may be checked that Q lies on the circumcircle. The lines DA', EB', FC' meet at the point Q' with co-ordinates (a2/(b – c), b2(c – a), c2/(a – b)) and it may be checked that Q' lies on the circumcircle. Interestingly Q' is the centre of the nine point rectangular hyperbola.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 4
(3.8)
(3.9) (3.10)
5
Article: CJB/97/2010
How the Excentres create Points on the Circumcircle Christopher Bradley Abstract: The circles through pairs of vertices of a triangle and the excentres opposite the third vertex have centres lying on the circumcircle and pass through the incentre of the triangle. The triangles with these centres as vertices exhibit properties that are described.
P
R I3
W D A
O3
V P3 I2 P2 O2
G
K J
E
Q
I
C
B S F P1
O1 T
U I1
Fig. 1 The two sets of excentres I1, I2, I3 and U, V, W 1. Introduction The best way of proceeding is to describe the various stages whereby Fig. 1 is constructed mentioning properties that are exhibited as they emerge. We start with a triangle ABC, its incentre I and its triangle of excentres I1I2I3. The segments I2I3, I3I1, I1I2 have midpoints D, E, F, the circle ABCDEF being the nine-point 1
circle of triangle I1I2I3. Thus far we have the same configuration of Article 96 (CJB/2010/96). J is the incentre of triangle DEF and G its centroid. The deLongchamps point of DEF is proved in Article 96 to coincide with I the incentre of ABC. A nine-point rectangular hyperbola passes through points I, J, I1, I2, I3, U, V, W, G, as described in Article 95 (CJB/2010/95). What now follows appears only in this article. Circles I1BC, I2CA, I3AB are now drawn and their centres O1, O2, O3 are shown to lie on the circumcircle of ABC. These circles all pass through I. Interestingly the lines I2I3, EF and O2O3 are parallel as are similar sets of lines including FD and DE. In fact the figures EFO2O3, FDO3O1 and DEO1O2 are rectangles. Circles UBC, VCA, WAB are now drawn and their centres P1, P2, P3 are shown to lie on the circumcircle of ABC. These circles all pass through J. The well known circles I1I2AB, I2I3BC, I3I1CA centres F, D, E also appear in Fig. 1. From this point on the results mentioned are not proved, but are indicated by CABRI II plus. Lines O1P1, O2P2, O3P3 form a triangle RST. The midpoints of the sides of R, S, T are P1, P2, P3. The points R, S, T are in fact the excentres of triangle O1O2O3. The two triangles O1O2O3 and P1P2P3 with vertices all lying on circle ABCDEF now take on the same properties as ABC and DEF, but with excentres R, S, T rather than I1, I2, I3 This process could apparently be carried on ad infinitum. The lines RI1, SI2, TI3 are concurrent at Q, the incentre of triangle O1O2O3. It is also indicated that circles RSW, STU, TRV, UVW all pass through a point P. 2. The circles I1BC, I2CA, I3AB The co-ordinates of the excentres are I1(– a, b, c), I2(a, – b, c), I3(a, b, – c). straightforward to find the equations of the circles I1BC, I2CA, I3AB, which are I1BC: bcx2 – a2yz + b(c – b)zx + c(b – c)xy = 0, I2CA: cay2 – b2zx + c(a – c)xy + a(c – a)yz = 0, I3AB: abz2 – c2xy + a(b – a)yz + b(a – b)zx = 0.
It is now (2.1) (2.2) (2.3)
Now the centre of a conic with equation ux2 + vy2 + cz2 + 2fyz + 2gzx + 2hxy = 0 (2.4) is (vw – gv – hw – f2 + fg + hf, wu – hw – fu – g2 + gh + fg, uv – fu – gv –h2 + hf + gh) see Bradley [1], so from Equations (2.1) to (2.3) we may obtain the co-ordinates of the centres O1, O2, O3 of the three circles . They are O1(– a2, b(b + c), c(b + c)), O2(a(c + a), – b2, c(c + a)), O3 (a(a + b), b(a + b), – c2) (2.5) It may now be checked that O1, O2, O3 all lie on the circumcircle of ABC with equation a2yz + b2zx + c2xy = 0. (2.6) It may also be checked that all three circles pass through I(a, b, c), the incentre of triangle ABC. 3. The points D, E, F and some of their properties 2
The points D, E, F are the midpoints of I2I3, I3I1, I1I2 respectively so their co-ordinates are D(a2, bc – b2, bc – c2), E(ca – a2, b2, ca – c2), F(ab – a2, ab – b2, c2). 3.1) It may be checked that D, E. F lie on the circumcircle of ABC. (In fact the circle ABCDEF is the nine-point circle of triangle I1I2I3, D, E, F being the midpoints of the sides.) The points U, V, W are the excentres of triangle DEF and it follows from Section 2 that the circles UEF, VFD, WDE all pass through J, the incentre of triangle DEF and that the centres of the three circles P1, P2, P3 lie on the circumcircle of triangle DEF, which is the same as circle ABC. The co-ordinates of U, V, W, J, P1, P2, P3 are very complicated and we do not record them. The well known circles I1I2AB, I2I3BC, I3I1CA centres F, D, E also appear in Fig. 1. The circle BCI1I2 has equation bcx2 + a2yz + b(b + c)zx + c(b + c)xy = 0. (3.2) The equations of CAI3I1, ABI1I2 may be written down from Equation (3.2) by cyclic change of x, y, z and a, b, c. 4. Further properties of the key points The equations of O2O3, I2I3 are respectively bcx – c(a + c)y – b(a + b)z = 0 (4.1) and cy + bz = 0. (4.2) and these are parallel since they meet at (c – b, b, – c) a point on the line at infinity x + y + z = 0. The equation of EF is bcx + c(a – c)y + b(a – b)z = 0 (4.3) and this too passes through (c – b, b, – c) and hence is parallel to both O2O3 and I2I3. Similarly O3O1, I3I1, FD are parallel and O1O2, I1I2, DE are parallel. The displacement O3E is (a(b + c), – ab, – ac)/{(a + b + c)(a – b – c)} and the displacement EF is (a(b – c), ab, ac)/{(a + b – c)(a – b + c)}. Now if (f, g, h) and (u, v, w) are two displacements (with f + g + h = 0 and u + v + w = 0, of course) then these displacements are at right angles if, and only if, a2(hv + gw) + b2(fw + hu) + c2(gu + fv) = 0. (4.4) It may now be checked that O3E and EF are at right angles. It follows that O3EFO2 is a rectangle as are O1FDO3 and O2DEO1. Reference 1. C. J. Bradley, The Algebra of Geometry, Highperception, Bath, UK (2007). Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article: CJB/2010/98
Ex-points and their Sets of Intersecting Circles Christopher Bradley Abstract: The ex-symmedian points and the ex-points of an arbitrary point internal to the triangle are shown to lie on a hyperbola. The configuration involving ex-points of a given point and vertices of a given triangle is shown to produce a set of three circles having a point in common (similar to the Fermat point). The relationship between these points is obtained.
F
W
P2 To S
A
P3
To P2
L' N
R
M
E
X
O
K P
Q M' B C'
C
A' Y L
N'
V P1
To U
D To B'
T
B'
Fig. 1 U The configuration for the symmedian point K and an arbitrary point P
1
1. Introduction The best way to introduce the results in this article is to describe the construction of Fig. 1 pointing out what needs to be proved. But first let us define what is meant by an ex-point. A point P internal to a triangle ABC has barycentric (areal) co-ordinates (l, m, n) with ABC as triangle of reference, where l, m, n > 0. Given such an internal point then the three ex-points P1, P2, P3 are defined to have co-ordinates (– l, m, n), (l, – m, n), (l, m, – n). Thus the ex-symmedian points D, E, F have co-ordinates (– a2, b2, c2), (a2 – b2, c2), (a2, b2, – c2) respectively. It is well-known that the ex-symmedian points are constructed by drawing the tangents at A, B, C to the circumcircle Σ of ABC, and then the tangents at B and C meet at D, those at C and A meet at E and those at A and B meet at F. The symmedian point in Fig. 1 is labelled K. The three circles DBC, ECA, FAB meet at O, the circumcentre of circle ABC. This is proved in what follows. That these three circles meet is obvious since they do so at the Miquel point of A, B, C with respect to triangle DEF, but it is desirable to find the precise position of this Miquel point. P is now chosen as an arbitrary point internal to triangle ABC. The line AP is then drawn meeting Σ at L. Points M and N are similarly defined on BP, CP respectively. The line DL meets Σ again at L'. Points M', N' are similarly defined on EM, FN respectively. It turns out that P 1 = AL^BM'^CN', P2 = AL'^BM^CN' and P3= AL'^BM'^CN. This is actually proved in Article 13, [CJB/2010/13] but we think it best to prove this rather intriguing result again. There are a number of further interesting results that now follow: (i) EP2, FP3 meet at a point A' on BC, with B', C' similarly defined on CA, AB respectively and we find their co-ordinates. (ii) Circles P1BC, P2CA, P3AB meet at the Miquel point of A, B, C with respect to triangle P1P2P3. This point is labelled X in Fig. 1 and we find its co-ordinates. (iii)A hyperbola passes through P, K, D, E, F, P1, P2, P3 and a ninth point Q = DL^EM^FN and we obtain its equation. (iv) The points U = MN'^NM'^ AP, V = NL'^LN'^BP, W = LM'^ML'^CP lie on a line that we identify. (v) The points R = P2P3^MN^M'N', S = P3P1^ NL^N'L', T= P1P2^LM^L'M' also lie on the line UVW. 2. The circles DBC, ECA, FAB meet at O The equation of circle DBC is b2c2x2 – (b2 + c2 – a2)yz + b2(a2 – b2)zx – c2(c2 – a2)xy = 0.
2
(2.1)
The equations of circles ECA, FBC may be written down from Equation (2.1) by cyclic change of x, y, z and a, b, c. It may then be checked that these three circles all pass through the circumcentre O of triangle ABC. 3. The points L, M, N, L', M', N', Q, P1, P2, P3 Starting with P(l, m, n) we find the equation of AP to be ny = mz. This meets the circumcircle Σ at the point L with co-ordinates L(– a2mn/(b2n + c2m), m, n). Similarly BP, CP meet Σ at M, N respectively with co-ordinates M(l, – b2nl/(c2l + a2n), n) and N(l, m, – c2lm/(a2m + b2l). The line DL has equation (3.1) This meets Σ again at the point L' with co-ordinates L'(a2mn/(b2n – c2m), – m, n) Note that this differs from L simply by a change of sign of m (or n). We can now write down, by cyclic change, the co-ordinates of M' and N', which are M'(l, b2nl/(c2l – a2n), – n) and N'(– l, m, c2lm/(a2m – b2l)). Now EM has equation that may be derived from that of DL by cyclic change and then the generating point Q is determined as DL^EM. The result is that Q has co-ordinates
It may now be checked that Q also lies on the line FN. The equations of the lines AL', BM', CN' can now be determined and they are respectively (3.2) 4. The circles P1BC, P2CA, P3AB and the Miquel point X created by P
We now have enough to determine the co-ordinates of the points P1 = AL^ BM'^CN', P2 = AL'^BM^CN' and P3 = AL'^BM'^CN, and, as you will have guessed, these are the ex-points of P with co-ordinates P1(– l, m, n), P2(l, – m, n), P3(l, m, – n). The equation of the circle P1BC is (a2mn – b2nl – c2lm)x2 + a2l(m + n – l)yz + (a2mn – b2l(l – m) – c2lm)zx + (a2mn – b2nl – c2l(l – n))xy.
3
(4.1)
The equations of circles P2CA and P3AB may now be written down from equation (4.1) by cyclic change of x, y, z and a, b, c and l, m, n. It may now be checked that these three circles meet at the point X with co-ordinates (x, y, z), where x = l(m + n – l)/(b2nl + c2lm – a2mn), y = m(n + l – m)(c2lm + a2mn – b2nl), (4.2) 2 2 2 z = n(l + m – n)(a mn + b nl – c lm). These are the co-ordinates of the Miquel point for points A, B, C on the sides the triangle of excentres derived from the point P(l, m, n). As such it is an important formula. 5. The hyperbola through the ex-points of K and P The equation of the conic passing through the ex-points P1, P2, P3, D, E, F is (b4n2 – c4m2)x2 + (c4l2 – a4n2)y2 + (a4m2 – b4l2)z2 = 0.
(5.1)
It may be checked that this hyperbola also passes through the points P, K and Q making it a nine point hyperbola as in Articles 96 and 97. [CJB/2010/96 and CJB/2010/97]. 6. The points A', B', C' and the line UVW The equation of the line EP2 is (b2n – c2m)x + (a2n – c2l)y +(a2m – b2l)z = 0.
(6.1)
The equations of FP3 and DP1 may now be written down from Equation (6.1) by cyclic change of x, y, z and a, b, c and l, m, n. The lines EP2 and FP3 meet at the point A'(0, a2m – b2l, c2l – a2n), which lies on BC. The coordinates of B' and C' may now be written down by cyclic change and they lie respectively on CA and AB. It follows that AA', BB', CC' meet at a point Y on the circumcircle with coordinates Y(1/(b2n – c2m), 1/(c2l – a2n), 1/(a2m – b2l)). The point U = MN'^AP^NM' has co-ordinates (x, y, z), where x = – l2(b2n + c2m), y = a2m2n, z = a2mn2.
(6.2)
Points V = NL'^BP^LN' and W = LM'^CP^ML' have co-ordinates that may be written down from (6.2) by cyclic change of x, y, z and a, b, c. The equation of the line UVW may now be worked out and is 4
a2x/l2 + b2y/m2 + c2z/n2 = 0.
(6.3)
Next the point R = P2P3^MN^M'N' has co-ordinates (x, y, z), where x = l2(b2n – c2m), y = – a2m2n, z = a2mn2. (6.4) It may now be checked that R lies on the line UVW as do S = P3P1^NL^N'L' and T = P1P2 ^ LM^L'M'. Cabri II plus indicates that the line UVW is the polar of the point Q with respect to Σ. There is no difficulty about checking this with algebra software, but it would not provide any further interest because of the technical difficulty involved.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article: CJB/2010/99
A Triangle and its Image under a Half Turn Christopher Bradley Abstract A triangle ABC and its image DEF under a half turn results in circles BCD, CAE, ABF intersecting at a point P on circle DEF. A point Q on circle ABC is similarly defined. Also it is established that a conic passes through A, B, C, D, E, F, P, Q
F
E A
Q
S
P
D B
C
Fig. 1 A Triangle ABC its image DEF and intersecting circles such as ABF, BCD, CAE
1
1. Introduction Given triangle ABC the points D, E, F are the images of A, B, C respectively under a half-turn rotation about an arbitrary point S. It is found that circles ABF, BCD, CAE have a common point P that also lies on the circumcircle of triangle DEF. Similarly (in fact, by symmetry) circles DEC, EFA, FDB have a common point Q lying on the circumcircle of triangle ABC. We also prove that A, B, C, D, E, F, P, Q lie on a conic. Areal co-ordinates are used throughout with ABC as triangle of reference. 2. The points D, E, F and circles BCD, CAE, ABF We suppose that S has co-ordinates (l, m, n). Then the co-ordinates of D, E, F are as follows: D(2l – 1, 2m, 2n), E(2l, 2m – 1, 2n), F(2l, 2m, 2n – 1). (2.1) Here it is important to realize expressions (2.1) are correct if, and only if, l + m + n = 1.
(2.2)
The equation of the circle BCD may now be worked and is best stated in the form a2yz + b2zx + c2xy + {2/(1 – 2l)}(2a2mn + (2l – 1)(b2n + c2m))x(x + y + z) = 0.
(2.3)
The equations of the circles CAE and ABF follow from Equation (2.3) by cyclic change of x, y, z and a, b, c and l, m, n. The co-ordinates of the common point P of the three circles are very involved, but may easily be checked using an algebra computer software package. 3. The conic ABCDEF If there is a conic through these six points then since it passes through A, B, C it must have an equation of the form fyz + gzx + hxy = 0, (3.1) for some constants f, g, h. Substituting in the co-ordinates of D and E we find the conic ABCDE has equation l(1 – 2l)yz + m(1 – 2m)zx + n(1 – 2n)xy = 0. (3.2) Its symmetric form shows that it also passes through F. 4. The point Q
2
The equation of circle ABC is of course a2yz + b2zx + c2xy = 0. It meets the conic ABCDEF at the point Q with co-ordinates (x, y, z), where x = 1/(b2n(2n – 1) – c2m(2m – 1)), y = 1/(c2l(2l – 1) – a2n(2n – 1)), z = 1/(a2m(2m – 1) – b2l(2l – 1)).
(4.1)
(4.2)
If S is at the centroid, then this point is, of course, the Steiner point. 5. The circle DEC The equation of circle DEC is 2(b2n(2l + 2m – 1) + c2m(2m – 1))x2 + 2(a2n(2l + 2m – 1) + c2l(2l – 1))y2 – (a2(2l + 2m – 1)2 + 2c2l(1 – 2l))yz – (b2(2l + 2m – 1)2 + 2c2m(1 – 2m))zx + (2a2n(2l + 2m – 1) + 2b2n(2l + 2m – 1) – (2l(4m + 2n – 1) + (2n – 1)(2m – 1)))c2xy = 0. (5.1) The equations of circle EFA, FDB may now be written down by cyclic change of x, y, z and a, b, c and l, m, n. It may now be verified that Q lies on these circles. By symmetry it follows that P lies on the conic and on the circles BCD, CAE, ABF, and DEF.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL, BS8 2SP.
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Article: CJB/2010/100
Circles though a point in an Equilateral Triangle Christopher Bradley Abstract: Given three points D, E, F lying on the medians of an equilateral triangle ABC the conditions are determined for the circles BCD, CAE, ABF to have a common point.
A
D
P Q
G
B
E C
F
Fig. 1 Showing positions of D, E, F in order that the two sets of circles have a common point
1. Introduction The purpose of this short article is to show that if E, F are fixed on the medians BG, CG of an equilateral triangle ABC, then there are two positions of D on the median AG for which circles BCD, CAE, ABF have a common point P. It then holds that with D, E, F in either set of positions circles AEF, BFD, CDE also have a common point Q, see [1] and see Fig. 1. Cabri II plus indicates that what holds for an equilateral triangle also holds for any triangle and for any set of Cevians (not just the medians). The general cases, however, are technically very involved and presenting them would not really be helpful. In proving the present propositions areal co-ordinates are used throughout with ABC as triangle of reference. See Bradley [2, 3] for an account of these co-ordinates and how to use them. 2. The points D, E, F and the circles BCD, CAE, ABF Arbitrary points D, E, F lying on the medians AG, BG, CG of an equilateral triangle ABC and centroid G have co-ordinates D(1 – 2d, d, d), E(e, 1 – 2e, e), F(f, f, 1 – 2f), where d, e, f are some real numbers. The equation of circle BCD is yz + zx + xy + d(3d – 2)x(x + y + z)/(1 – 2d) = 0.
(2.1)
The equations of circles CAE, ABF may now be written down by cyclic change of x, y, z and d, e, f. 3. The common point P of intersection of the three circles The circles BCD and CAE have common chord (2e – 1)(3d – 2)dx + (e(3e – 2) – 2de(3e – 2))y = 0.
(3.1)
The equation of the common chord of circles CAE and ABF may be written down by cyclic change of x, y, z and d, e, f. These two common chords meet at a point P with co-ordinates (x, y, z), where x = (2d – 1)/{d(3d – 2)}, y = (2e – 1)/{e(3e – 2)}, z = (2f – 1)/{f(3f – 2)}. This point lies on all three circles if either d = (3(e + f) – 3ef – 2)/{3(e + f – 1)},
(3.2)
(3.3)
or d = (3ef – e – f)/(6ef – 3e – 3f + 1).
(3.4)
This shows that if E and F are fixed, then there are two positions of D for which the three circles have a common point. In Article 19, see [1], we call this situation a case of circular perspective between circles ABC and DEF (the triangles ABC and DEF in this case also being in perspective). And we prove in that article that this relationship is symmetric, and hence circle AEF, BFD, CDE also have a common point Q. References 1. 2. 3.
C. J. Bradley, Article CJB/2010/19 of this series; C.J.Bradley, Challenges in Geometry, Oxford (2005); C.J.Bradley, The Algebra of Geometry, Highperception, Bath (2007).
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
Article: CJB/2010/101
Intersections at the vertices of the Second Brocard Triangle Christopher Bradley Abstract: If D, E, F are the ex-symmedian points it is proved that circles BCD, CAE, ABF intersect the Brocard circle at the circumcentre of ABC and at the vertices of the second Brocard triangle. A case of circular perspective is thereby determined.
F
A
E W K
O
U
V B
C
P
D
Fig. 1 1
Circles intersecting on the Brocard circle 1. Introduction In Fig.1 O is the circumcentre of triangle ABC and K is the symmedian point. Points D, E, F are the ex-symmedian points (where the tangents at A, B, C to the circumcircle intersect). The circle on OK as diameter is the Brocard circle. In this short article we prove that circle BDC meets the Brocard circle at O and at U, the first vertex of the second Brocard tiangle, which also lies on line AKD. V and W, the second and third vertices of the second Brocard triangle are determined in similar fashion by circles CAE and ABF respectively. It is also shown that circles AVW, BWU, CUV pass through the same point P on the circumcircle. 2. The Brocard circle and the ex-symmedian points The Brocard circle passes through O, K and the two Brocard points and has equation b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0.
(2.1)
The three ex-symmedian points have co-ordinates D(– a2, b2, c2), E(a2, – b2, c2) and F(a2, b2, – c2). 3. Circles BCD, CAE, ABF The equation of the circle BCD is b2c2x2 – a2(b2 + c2 – a2)yz + b2(a2 – b2) zx + c2(a2 – c2)xy = 0.
(3.1)
The equations of circles CAE, ABF may now be written down from Equation (3.1) by cyclic change of x, y, z and a, b, c. 4. The Points of intersection Circle BCD with equation (3.1) and the Brocard circle with equation (2.1) meet at O the circumcentre of triangle ABC and at the point U(b2 + c2 – a2, b2, c2). U is the first point of the second Brocard triangle (which is in perspective with the first Brocard triangle with perspector the centroid). The co-ordinates of the points V, W the intersections of circles CAE and ABF other than O with the Brocard circle may now be written down from those of U by cyclic change of x, y, z and a, b, c. They are, of course, the second and third vertices of the second Brocard triangle. From the co-ordinates of U, V, W, K, D, E, F it is also clear that AKUD, BKVE, CKWF are straight lines. 5. The intersection of circles AVW, BWU, CUV 2
The fact that circles BCU, CAV, ABW meet at the circumcentre O means that triangles ABC and UVW are in circular perspective (see [1], Article CJB/2010/19) which is a symmetric relationship means that circles AVW, BWU, CUV also have a common point P. The coordinates of P are (a2/(b2 – c2), b2/(c2 – a2), c2/(a2 – b2)). It may be checked that P lies on the circumcircle of triangle ABC. Reference 1. C. J. Bradley, Article 19 of this series, Article: CJB/2010/19.
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP.
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4
F
A
W K
O
U
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Article: CJB/2010/102
Two Triangles, their Ex-Symmedians and four Conics Christopher Bradley
To Y
C'
W D'
13
A
F
12
cd bd
B'
E ce
23 To X X
J
U
bf
K
32
G B
C
ae
af
E'
D 21
31
F'
To Z A'
V
Fig. 1 Two triangles, four conics and a pair of parallel lines
1
Abstract: The two triangles are in perspective, vertex the centroid of one of them. The six exsymmedian points lie on a conic, the tangents at the vertex circumscribe the circumcircle and also the intersections of various lines produce six points lying three by three on two parallel lines. Two other conics of interest are also created. 1. Introduction A triangle ABC, centroid G and circumcircle Σ is given. AG, BG, CG are drawn to meet Σ again at D, E, F respectively. Tangents at A, B, C to Σ are drawn, pairs of which meet at the exsymmedian points A', B', C'. Similarly tangents to Σ are drawn at D, E, F, pairs of which meet at the ex-symmedian points D', E', F'. Lines B'C' and EF meet at U, lines C'A' and FD meet at V and lines A'B' and DE meet at W. We prove UVW is a straight line. Lines BC and E'F' meet at X, lines CA and F'D' meet at Y and lines AB and D'E' meet at Z. We prove XYZ is a straight line parallel to UVW. The tangents at A and E meet at the point 12, the tangents at A and F meet at 13, the tangents at B and D meet at the point 21, the tangents at B and F meet at 23, the tangents at C and D meet at 31 and the tangents at C and E meet at 23. We prove that 13, 12, 32, 31, 21, 23 lie on a conic. The sides of the two triangles ABC and DEF meet at cd = AB^EF, bd = CA^EF, bf = AC^DE, af = BC^DE, ae = BC^FD, ce = AB^FD. We prove that these six points also lie on a conic. Fig. 1 illustrates all these properties and in addition shows many lines passing through G, for example 13 31. When co-ordinates have been established the equations of such lines may easily be obtained showing that they do in fact pass through G. Such additional properties are left to the reader, as this article is quite long enough as it is! We use areal co-ordinates throughout with ABC as triangle of reference. 2. Points D, E, F and the ex-symmedians of the two triangles The line AG has equation y = z and meets the circumcircle Σ with equation a2yz + b2zx + c2xy = 0 at the point D with co-ordinates D(– a2, b2 + c2, b2 + c2).
(2.1)
Similarly the co-ordinates of E, F where BG, CG meet Σ are E(c2 + a2 , – b2, c2 + a2 ), F(a2 + b2 , a2 + b2, – c2).
2
The ex-symmedians A', B', C' of triangle ABC are well-known to have co-ordinates A'(– a2, b2, c2), B'(a2, – b2, c2), C'(a2, b2, – c2). The tangent to Σ at D has equation (b2 + c2)2x + a2b2y + a2c2z = 0.
(2.2)
The equations of the tangents to Σ at E, F may be written down from Equation (2.2) by cyclic change of x, y, z and a, b, c. The ex-symmedian point D' of triangle DEF is the intersection of the tangents at E and F and therefore has co-ordinates (x, y, z), where x = 2b2c2 + a2(a2 + b2 + c2), y = – b2(a2 + b2 – c2), (2.3) 2 2 2 2 z = – c (c + a – b ). The co-ordinates of E' and F' may now be written down from those of D' by cyclic change of x, y, z and a, b, c. 3. The conic and lines through the ex-symmedian points The equation of this conic may now be worked out and is a2b2c2((b2 + c2) x2 + (c2 + a2) y2 + (a2 + b2) z2) + (a4( b2 + c2) + b4(c2 + a2) + c4(a2 + b2))(a2yz + b2zx + c2xy) = 0.
(3.1)
It is well-known that AA', BB', CC' meet at the symmedian point K of triangle ABC. The equation of the line DD' is (b4 – c4)x + b2(2c2 + a2)y – c2(a2 + 2b2)z = 0. (3.2) The equations of lines EE', FF' may be written down from Equation (3.2) by cyclic change of x, y, z and a, b, c. These three lines concur at the symmedian point J of triangle DEF with coordinates J(2b2c2 + a2(b2 + c2 – a2), 2c2a2 + b2(c2 + a2 – b2), 2a2b2 + c2(a2 + b2 – c2)), a point lying on the line through the circumcentre O and the isotomic conjugate of K. 4. The parallel lines UVW and XYZ The equation of EF is – a2x + (c2 + a2)y + (a2 + b2z) = 0,
(4.1)
and the equations of FD, DE follow by cyclic change of x, y, z and a, b, c. The equation of B'C' is c2y + b2z = 0,
(4.2) 3
and the equations of C'A' and A'B' follow by cyclic change of x, y, z and a, b, c. The point U is the intersection of EF and B'C' and so has co-ordinates U( b2 – c2, b2, – c2). The point V is the intersection of FD and C'A' and has co-ordinates V(– a2, c2 – a2, c2). The point W is the intersection of DE and A'B' and has co-ordinates W(a2, – b2, a2 – b2). It may now be checked that U, V, W all lie on the line with equation (b2 + c2 – a2)x + (c2 + a2 – b2)y + (a2 + b2 – c2)z = 0.
(4.3)
The line E'F' has equation c2a2x + b2c2y + (a2 + b2)2z = 0, (4.4) 2 2 and this meets BC at the point X with co-ordinates (0, c , – b ). Similarly F'D' meets CA at the point Y with co-ordinates (– c2, 0, a2) and D'E' meets AB at the point Z with co-ordinates (b2, – a2, 0). Clearly X, Y, Z all lie on the line with equation a2x + b2y + c2z = 0. (4.5) Note that UVW and XYZ are parallel since they meet at the point with co-ordinates (b2 – c2, c2 – a2, a2 – b2), which lies on the line at infinity with equation x + y + z = 0. Note that the Euler line of triangle ABC is perpendicular to both UVW and XYZ. Cabri II plus indicates that the line UVW is the polar of a point on the Euler line OH beyond H. 5. The points 21, 32, 13, 12, 23, 31 and the conic on which they lie The tangent at D has Equation (2.2) and that at B is c2x + a2z = 0. These two tangents meet at the point 21 with co-ordinates 21(– a2, b2 + 2c2, c2). The tangents C and E meet at the point 32 with co-ordinates 32(a2, – b2, c2 + 2a2). The tangents at A and F meet at the point 13 with co-ordinates 13(a2 + 2b2, b2, – c2). The tangents at D and C meet at the point 31 with co-ordinates 31(– a2, b2, 2b2 + c2). The tangents at E and A meet at the point 12 with co-ordinates (2c2 + a2, – b2, c2). The tangents at F and B meet at the point 23 with co-ordinates 23(a2, 2a2 + b2, – c2). These six points may now be shown to lie on the conic with equation a2b2c2(a2(b2 + c2)x2 + b2(c2 + a2)y2 + c2(a2 + b2)z2) + (a6(b2 + c2) + b6(c2 + a2) + c6(a2 + b2) + 4a2b2c2(a2 + b2 + c2))(a2yz + b2zx + c2xy) = 0.
(5.1)
6. The points ae, bf, cd, ce, af, bd and the conic on which they lie These are the six internal points of intersection of the triangles ABC and DEF. The point ae = FD^BC has co-ordinates (0, a2 + b2, b2). The point bf = DE^CA has co-ordinates (c2, 0, b2 + c2). The point cd = EF^AB has co-ordinates (c2 + a2, a2, 0). The point ce= FD^AB has co-ordinates (b2, b2 + c2, 0). The point af = DE^BC has co-ordinates (0, c2, c2 + a2). The point bd = EF^CA 4
has co-ordinates (a2 + b2, 0, a2). These six points may now be shown to lie on the conic with equation a2(b2 + c2)x2 + b2(c2 + a2)y2 + c2(a2 + b2)z2 – (2b2c2 + a2(a2 + b2 + c2))yz – (2c2a2 + b2(a2 + b2 + c2)zx – (2a2b2 + c2(a2 + b2 + c2))xy = 0. (6.1) It may be asked which of the conics remains if a perspective point other than G is used to create Triangle DEF. The answer is that Cabri II plus indicates all four conics remain. It is, however, beyond my computing power to prove it. G was chosen because the working was manageable.
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/103
The Altitudes Create Four Circles, Four Conics and a Polar line
cf
Christopher Bradley 33'
To cf 12
2'3' To 33' 3'1
23' B
A 12' cd F 31 d c bd E ce O H b f 2'3 bf e ae a af C 1'2'
11' ad
D 31' To 3'1'
1'2
23 To 22'
3'1' be
Fig. 1 The four circles, four conics and the polar line of H
22'
Abstract: If the altitudes of a triangle ABC meet the circumcircle at D, E, F, then the six interior intersections of the two triangles and the tangents at the above six points produce a 1
configuration in which four circles and four conics feature. An analysis is given in which the polar of the orthocentre plays a major role. 1. Introduction The best introduction is a description of how Fig. 1 was constructed, introducing the conics and circles as they emerge. In the following sections analytic verifications are made of the results, using areal co-ordinates with ABC as triangle of reference. It is true, of course, that many of the results follow immediately from simple synthetic arguments, but the equations of the conics and circles are significant in their own right. The start, of course, is a triangle ABC and its circumcircle Σ. The altitudes are then drawn meeting Σ at D, E, F. We now have two triangles ABC and DEF. Points are now constructed from the following lines: the sides AB, BC, CA, DE, EF, FD and the tangents to Σ at A. B, C, D,. E, F. The internal points of intersection of the two triangles are cd = AB^EF, ce = AB^FD, ae = BC^FD, af = BC^DE, bf = CA^DE, bd = CA^EF. The external points of intersection are ad = BC^EF, be = CA^FD, cf = AB^DE. It is proved that cd, ce, ae, af, be, cf lie on a conic Points on the tangents at A, B, C carry labels 1, 2, 3 respectively and tangents at D, E, F carry labels 1', 2', 3’. So 11' is the intersection of the tangents at A and D, 22' is the intersection of the tangents at B and E, 33' is the intersection of the tangents at C and F. It is then the case that 11', 22', 33', ad, be, cf lie on a line, which we prove to be the polar of the orthocentre H with respect to Σ. If one now thinks of the four points ABDC in that order, we may consider AD and BC to be two perpendicular diagonals of a cyclic quadrilateral. It follows then from a well-known theorem that the tangents at these four points meet in pairs in another circle. It follows that 12, 31, 1'2 and 31' are concyclic. Similarly 23, 12, 2'3, 12' are concyclic as are 31, 23, 3'1, 23'. The ex-symmedian points of triangles ABC and DEF, namely 12, 23, 31, 1'2', 2'3', 3'1' are proved to lie on a conic. The points 12', 23', 31', 1'2, 2'3, 3'1 are also proved to lie on a conic. If the altitude AD meets EF at d and BC at a, with e, b, f, c similarly defined, then finally it is shown that these six points lie on a conic. 2. Points D, E, F, the sides EF, FD, DE and the tangents to Σ at A, B, C, D, E, F We take the orthocentre H of triangle ABC to have co-ordinates (u, v, w), so that the equation of the circumcircle Σ is u(v + w)yz + v(w + u)zx + w(u + v)xy = 0. (2.1) 2
The equation of AH is wy = vz and this line meets Σ at the point D(– u(v + w), v(2u + v + w), w(2u + v + w)). Similarly E has co-ordinates E(u(2v + w + u), – v(w + u), w(2v + w + u)) and F has co-ordinates F(u(2w + u + v), v(2w + u + v), – w(u + v)). We are now able to find the equations of the sides of triangle DEF. The equation of EF is vw(v + w)x – wu(u + 2v + w)y – uv(u + v + 2w)z = 0. (2.2) The equations of FD and DE follow from Equation (2.2) by cyclic change of x, y, z and u, v, w. The tangent at A to Σ has equation w(u + v)y + v(w + u)z = 0.
(2.3)
The tangents at B and C to Σ follow from Equation (2.3) by cyclic change of x, y, z and u, v, w. The tangent at D to Σ has equation vw(2u + v + w)2x + wu(v + w)(w + u) y + uv(v + w)(u + v)z = 0. (2.4) The tangents at E and F to Σ follow from Equation (2.4) by cyclic change of x, y, z and u, v, w. 3. Points 12, 23, 31, 1'2', 2'3', 3'1' and the conic through these six points The tangents at A and B to Σ meet at the point 12 whose co-ordinates are (u(v + w), v(w + u), – w(u + v)). Similarly point 23 has co-ordinates (– u(v + w), v(w + u), w(u + v)) and the point 31 has co-ordinates (u(v + w), – v(w + u), w(u + v)). Tangents at D and E to Σ meet at the point 1'2' whose co-ordinates are (– uv(v + w), – uv(w + u), w(u2 + v2 + w2 + 3uv + 2vw + 2wu)). Similarly point 2'3' has co-ordinates (u(u2 + v2 + w2 + 2uv + 3vw + 2wu), – vw(w + u), – vw(u + v)) and 3'1' has co-ordinates (– wu(v + w), v(u2 + v2 + w2 + 2uv + 2vw + 3wu), – wu(u + v)). We are now in position to work out the equation of the conic through these six points, which are the ex-symmedian points of the two triangles. It is px2 + qy2 + rz2 + 2fyz + 2gzx + 2hxy = 0, (3.1) where p = 2v2w2(u + v)(v + w)(w + u)(2u + v + w), q = 2w2u2(u + v)(v + w)(w + u)(u + 2v + w), r = 2u2v2(u + v)(v + w)(w + u)(u + v + 2w), f = u2vw(v + w)k, g = v2wu(w + u)k, h = w2uv(u + v)k, (3.2) 3
and where k = 2(u3 + v3 + w3) + 5(uv2 + uw2 + vw2 + vu2 + wu2 + wv2) + 12uvw.
(3.3)
4. The points 11', 22', 33' and the line joining them Tangents at A and D to Σ meet at the point 11' whose co-ordinates are (u(v + w)(w – v), – v(w + u)(2u + v + w), w(u + v)(2u + v + w)).
(4.1)
Points 22' and 33' have co-ordinates that may be written down from (4.1) by cyclic change of x, y, z and u, v, w. These three points lie on the line with equation vw(2u + v + w)x + wu(2v + w + u)y + uv(2w + u + v)z = 0.
(4.2)
It may be checked that this is the polar of the orthocentre H with respect to the circumcircle Σ. 5. The points a, b, c, d, e, f and the conic through these six points The points a, b, c are the points where the altitudes meet the opposite sides and so have coordinates (0, v, w), (u, 0, w), (u, v, 0) respectively. The line EF has Equation (2.2) and AD has equation wy = vz. These lines meet at the point d with co-ordinates d(u(2u + 3v + 3w), v(v + w), w(v + w)). Similarly the points e and f have coordinates e(u(w + u), v(2v + 3w + 3u), w(w + u)) and f(u(u + v), v(u + v), w(2w + 3u + 3v)). We may now find the equation of the conic through the points a, b, c, d, e, f, which is equation (3.1), where now p = 2v2w2(v + w), q = 2w2u2(w + u), r = 2u2v2(u + v), (5.1) 2 f = – u vw(2u + v + w), g = – v2wu(u + 2v + w), h = – w2uv(u + v + 2w). 6. The points ad, be, cf lie on UVW, the polar of H The point ad = BC^EF. The line EF has equation (2.2) so ad has co-ordinates (0, – v(u + v + 2w), w(u + 2v + w)). Points be, cf have co-ordinates that may be obtained from those of ad by cyclic change of x, y, z and u, v, w. It may now be checked that these three points all lie on the line UVW with Equation (4.2) and which is the polar of H with respect to Σ. 4
7. Points 12, 31, 1'2, 31' lie on a circle Lines AD and BC are perpendicular diagonals of the cyclic quadrilateral ABDC and so the tangents at A, B, D, C form the circle 12, 31, 1'2, 31'. The tangents at A and B to Σ meet at the point 12 whose co-ordinates are (u(v + w), v(w + u), – w(u + v)). Similarly point 31 has co-ordinates (u(v + w), – v(w + u), w(u + v)). The tangent at D to Σ has equation (2u + v + w)2vwx + (v + w)(w + u)wuy + (u + v)(v + w)uvz = 0,
(7.1)
and the tangent at B has equation (u + v)wx + (v + w)uz = 0.
(7.2)
These two lines meet at the point 1'2 with co-ordinates (– u(v + w), v(3u + 2v + w), w(u + v)). Similarly 31' has co-ordinates (– u(v + w), v(w + u), w(3u + v + 2w)). It may now be checked that the four points lie on the circle with equation vw(u + v)(v + w)(w + u)(u + v + w)x2 + w2u(u + v)(v + w)(w + u)y2 + v2u(u + v)(v + w)(w + u)z2 + u(v + w)(((u2 + u(v + w))(v2 + 4vw + w2)) + vw(v2 + w2))yz + u2(v2 + 6vw + w2) + u(v + w)(v2 + 6vw + w2) + vw(v + w)2)((u + v)wxy + v(w + u)zx) = 0. (7.3) In similar manner it may be shown that points 23, 12, 2'3, 12' lie on a circle, as do points 31, 23, 3'1, 23'. So together with the circumcircle we have four circles in the configuration.
8. The points cd, ae, bf, bd, ce, af and the conic through these points The point cd = AB^EF. The line EF has Equation (2.2) and AB has equation z = 0. The coordinates of CD are found to be cd(u(u + 2v + w), v(v + w), 0). Similarly ae has co-ordinates ae(0, v(u + v + 2w), w(w + u)) and bf has co-ordinates (u(u + v), 0, w(2u + v + w)). Now bd = CA^EF and the co-ordinates of bd are found to be bd(u(u + v + 2w), 0, w(v + w)). Similarly ce has co-ordinates ce(u(w + u), v(2u + v + w), 0) and af has co-ordinates af(0, v(u + v), w(u + 2v + w)). It may now be checked that these six points lie on the conic with Equation (3.1), where now p = v2w2(v + w)(2u + v + w), q = w2u2(w + u)(u + 2v + w), r = u2v2(u + v)(u + v + 2w), (8.1) 2 2 2 2 f = – u vw(u + v + w + 2uv + 3vw + 2wu), 5
g = – v2wu(u2 + v2 + w2 + 2uv + 2vw + 3wu), h = – w2uv(u2 + v2 + w2 +3uv + 2vw + 2wu). It may be added that when any pair of triangles with a common circumcircle are in perspective then Cabri indicates that four conics exist of the type that are generated in this paper. The existence of four circles is, however, limited to when the vertex of perspective is the orthocentre of one of the triangles.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article : 2010/CJB/104
How any Six Points on a Circle create Two Conics Christopher Bradley Abstract: Given three chords of a circle, perpendiculars to those chords from their end points create three pairs of parallel lines. Of their fifteen points of intersection three lie on the line at infinity and the other twelve lie six by six on two conics. Note that the six points must lie on a circle and not a general conic.
ea
A B
bd
ab fb
ad
be
F
O cd
ef E ce
C cf
ac
D
df Fig. 1 Six points on a circle and the two conics arising
1
1. Introduction Six points A, B, C, D, E lie on a circle and chords BC, DE, FA are selected. Parallel lines through B and C are drawn perpendicular to BC, similarly for D and E and the chord DE and again for F and A and the chord FA. These three pairs of parallel lines meet at three points on the line at infinity and twelve points that lie six and six on a pair of conics. To be precise, using the notation xy to denote the intersection of lines through X and Y, the following six lie on a conic Σ: ab, ad, cd, cf, ef, be and the following six lie on a conic Γ: ea, bd, ac, df, ce, fb. Cabri II plus indicates that if the initial points lie on a conic other than a circle, then Σ still exists, but not Γ. However Γ can be made to exist by adjustment of the positions of the points, but that is not a matter of significance. In the following sections we prove the two results using Cartesian co-ordinates with the circle ABCDEF having equation x2 + y2 = 1. 2. The conic Σ through ab, ad, cd, cf, ef, be We take the co-ordinates of A to be ((1 – a2)/(1 + a2), 2a/(1 + a2)). The co-ordinates of B, C, D, E, F are the same but with parameters b, c, d, e, f respectively. The chord BC has equation (1 – bc)x + (b + c)y = (1 + bc). The line perpendicular to BC through B has equation (b + c)x – (1 – bc)y = c – b, and the line perpendicular to BC through C has equation (b + c)x – (1 – bc)y = b – c.
(2.1)
(2.2) (2.3)
The other two pairs of parallel lines have similar equations with d, e and f, a replacing b, c in turn. Note now that a point such as ab is the intersection of the lines with equation (f + a)x – (1 – af)y = f – a and Equation (2.2) and therefore has co-ordinates (x, y), where x = – (abc – abf + acf – bcf – a + b – c + f)/(abc – abf – acf + bcf – a + b + c – f), y = 2(ac – bf)/(abc – abf – acf + bcf – a + b + c – f ).
2
(2.4) (2.5) (2.6)
Similarly the co-ordinates of ad, cd, cf, ef, be may be written down. The equation of the conic Σ containing these six points is ux2 + vy2 + 2wxy + k = 0, where u = abd – abe – acd – 3ace + 3fbd + fbe + fcd – fce, v = – abcde + abcdf – abcef + abfde – acfde + bcdef + abc – abf + acf + ade – adf + aef – bcd + bce – bcf – bde + cde – def – a + b – c + d – e + f. w = – abce + abdf – acde – acef + bcdf + bdef + ac + ae – bd – bf + ce – df. k = (a – f)(b – c)(e – d).
(2.7) (2.8) (2.9) (2.10) (2.11)
3. The conic Γ through ea, bd, ac, df, ce, fb There is a problem! The conic concerned is of the form of Equation (2.7), but the expressions obtained by DERIVE for each of u, v, w contain of the order of 100 terms. All I can reasonably do is to offer anyone who writes to me a paper copy of their values, alongside k = (a – f)(b – c)(e – d)(abcd – abde + abef – adcf – becf + cdef + ab – ad – be +cd – cf + ef), which itself would have 96 terms if multiplied out, but unfortunately u, v, w do not factorize. (3.1)
Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP
3
Article CJB/2010/105
External Squares on the Sides of a Triangle Christopher Bradley Abstract: If squares are drawn externally on the sides of a triangle the configuration exhibits two points similar to the Fermat points. Also the six intersections of various lines define a circle.
aa
ac A'' ab
F
A' B'
A
W o
V O
Q
C''
P
cb
C
B
bc
E
U
bb
B''
ba
.
D
C'
ca
Fig. 1 Squares drawn externally on the sides of a Triangle
1
cc
1. Introduction In Fig. 1 squares BCC'B'', CAA'C'', ABB'A'' are drawn externally on the sides BC, CA, AB respectively of a triangle ABC. The sides of the triangle and the sides of the squares parallel to them are extended. The latter sides intersect at points aa, bb, cc. The side AB meets side C'B'' at point ba and the side AC meets side C'B'' at the point ca. Points ab, cb on side A'C'' and points ac, bc on side B'A'' are similarly defined. Points U, V, W are the centres of the squares BCC'B'', CAA'C'', ABB'A'' and points D, E, F are the midpoints of B''C', C''A', A''B' respectively. The following results now hold: (i) Lines AD, BE, CF are concurrent at a point Q; (ii) Lines AU, BV, CW are concurrent at a point P. (iii) The points ab, cb, ca, ba, bc, ac lie on a circle s centre o; In the following sections these results are proved using areal co-ordinates with ABC as triangle of reference. 2. Co-ordinates of key points As we are dealing with areal co-ordinates, we can use elementary expressions for the areas of points such as B''. Denoting the side lengths by a, b, c as usual and the area of triangle ABC by k, we find co-ordinates of B'' to be B''( – ½ a2, k + ¼ (a2 + b2 – c2), ¼ (c2 + a2 – b2)) and those of C' to be C'( – ½ a2, ¼ (a2 + b2 – c2), ¼ ( c2 + a2 – b2)). The equation of the line B''C' may now be obtained and is (a2 + 2k)x + a2(y + z) = 0. (2.1) The co-ordinates of C'' and A'' may now be obtained from those of B'' by cyclic change of x, y, z and a, b, c. Similarly those of A' and B' may be obtained from those of C'. And the lines C’’A’ and A''B' may also be obtained from Equation (2.1) by cyclic change. 3. The points D, E, F and the concurrency of AD, BE, CF at Q The point D is the midpoint of B''C' and so has co-ordinates (– ½ a2, ¼(a2 + b2 – c2 + 2k), ¼(a2 – b2 + c2 + 2k)). Points E, F may be obtained from those of D by cyclic change of x, y, z and a, b, c). It is now immediate that AD, BE, CF are concurrent at a point Q with co-ordinates (x, y, z), where x = 1/(b2 + c2 – a2 + 2k), y = 1/(c2 + a2 – b2 + 2k), (3.1) 2 2 2 z = 1/(a + b – c + 2k). 2
4. The points U, V, W and the concurrency of AU, BV, CW at P The point U is the centre of the square and therefore has co-ordinates one-quarter of those at the vertices B, B, C', B'' (all normalized to k) and these are (x, y, z), where x = – ¼a2, y = (1/8)( a2 + b2 – c2 + 4k) (4.1) 2 2 2 z = (1/8)(a – b + c + 4k). The co-ordinates of points V, W may now be obtained from Equations (4.1) by cyclic change of x, y, z and a, b, c. It is now immediate that AU, BV, CW meet at the point P with co-ordinates (x, y, z) where x = 1/(b2 + c2 – a2 + 4k), y = 1/(c2 + a2 – b2 + 4k), (4.2) 2 2 2 z = 1/(a + b – c + 4k). Note that the line PQ has equation (b2 + c2 – a2 + 2k)(b2 + c2 – a2 + 4k)(b2 – c2) x + ... + ... = 0. (4.3) 5. The points cc, aa, bb and ba, ca, cb, ab, ac, ca and the circle s The point cc = B''C'^C''A' has co-ordinates cc(– a2, – b2, a2 + b2 + 2k). The co-ordinates of aa, bb may be obtained from those of cc by cyclic change of x, y, z and a, b, c. The point ba = AB^B''C' and has co-ordinates (– a2, a2 + 2k, 0). The co-ordinates of cb, ac may be obtained from those of ba by cyclic change of x, y, z and a, b, c. The point ca = AC^B''C' and has co-ordinates (– a2, 0, a2 + 2k). The co-ordinates of ab, bc may be obtained from those of ca by cyclic change of x, y, z and a, b, c. It may now be shown that the six points ba, ca, cb, ab, ac, ca lie on a circle s with equation b2c2(a2 + 2k)x2 + ... + ... + 2(b2c2 + (b2 + c2)k + 2k2)yz + ... + ... = 0. (5.1) The centre of this circle turns out to be o(– a2(b2 + c2 + 2k) + b4 + c4 + 2k(b2 + c2), ..., ...). (In the above the dots denote cyclic permutations of a, b, c). Flat 4 Terrill Court 12-14, Apsley Road, BRISTOL BS8 2SP. 3
4
Article: CJB/2010/106
On Perpendiculars at the corners of a Cyclic Quadrilateral Christopher Bradley Abstract: If perpendiculars to the sides are drawn at the vertices of a cyclic quadrilateral then a configuration is created that consists of two circles coaxal with the first circle, each containing quadrilaterals similar to each other. Their external diagonal points also lie three by three on two collinear lines. Two other circles also emerge.
F
ab
F' O''
A
B' ad
C'
G'
dc
X D'
U
O
cd
da D Y
V G
bc C
ba
G'' B A'
O'
F'' cb
Fig.1 Perpendiculars at the vertices of a cyclic quadrilateral create four other circles 1
1. Introduction A cyclic quadrilateral ABCD is given inscribed in a circle S, centre O. Point A' is the point on S diametrically opposite A and points B', C', D' are similarly defined. A' may also be thought of as the intersection of the perpendicular to AB through B and the perpendicular to AD through D. The critical step in building the configuration of Fig. 1 is the creation of eight new points as the intersection of chords of the eight points so far defined on S. These are as follows (i) ad = AD'^DC', (ii) ba = BA'^AD', (iii) cb = CB'^BA', (iv) dc = DC'^CB', (v) da = DA'^AB', (vi) ab = AB'^BC', (vii)bc = BC'^ CD', (viii) cd = CD'^DA'. The following results now hold: (i) points ad, ba, cb, dc are concyclic, (ii) points da, ab, bc, cd are concyclic, (iii) points ab, da, dc, ad are concyclic, and (iv) points ba, cb, cd, bc are concyclic. Define the circle consisting of points (i) as S', centre O' abd the circle consisting of points (ii) as S'', centre O'', then (v) S, S', S'' are coaxal, with common chord XY (vi) O'O = OO''. Further, if F = AB^CD, F' = ba ab^dc cb and F'' = ab bc^cd da then (vii) F, F', F'' are collinear and if G = BC^AD, G' = cb ba^ad dc and G'' = da ab^bcc d then (viii) G, G', G'' are collinear. Finally (ix) XY, BA', CD' are concurrent at U and (x) XY, AB', DC' are concurrent at V. In the following sections we prove these results using Cartesian co-ordinates with circle S having equation x2 + y2 = 1, with O as origin. 2. Setting the scene Let A have co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2 )) and let B, C, D have similar co-ordinates, but with parameters b, c, d respectively rather than a. The perpendicular to AB through A has equation (a + b)x – (1 – ab)y = b – a. (2.1) This line meets S at the point B' with co-ordinates ((b2 – 1)/(b2 + 1), – 2b/(1 + b2)). C', D', A' have similar co-ordinates, but with parameters c, d, a respectively, rather than b. Note, of course, that these points on S lie diametrically opposite those of B, C, D, A. The equations of the eight chords we want are AB': (a + b)x + (ab – 1)y + a – b = 0; AD': (a + d)x + (ad – 1)y + a – d = 0; BC': (b + c)x + (bc – 1)y + b – c = 0; BA': (b + a)x + (ba – 1)y + b – a = 0; CD': (c + d)x + (cd – 1)y + c – d = 0; CB': (c + b)x + (cb – 1)y + c – b = 0; 2
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7)
DA': DC':
(d + a)x + (da – 1)y + d – a = 0; (d + c)x + (dc – 1)y + d – c = 0.
We can now compute the co-ordinates of ad = AD'^DC' and these are (x, y), where x = (2acd – (a + c)(d2 + 1) + 2d)/((a – c)(d2 + 1)), y = (2(ac – d2))/((c – a)(d2 + 1)).
(2.8) (2.9)
(2.10) (2.11)
The co-ordinates of ba = BA'^AD', cb = CB'^BA', dc = DC'^CB' may be written down from Equation (2.11) by cyclic change of a, b, c, d. We can also compute the co-ordinates of ab = AB'^BC' and these are (x, y), where x = ((b2 + 1)(a + c) – 2abc – 2b)/((c – a)(b2 + 1)), y = (2(ac – b2))/((c – a)(b2 + 1)).
(2.12) (2.13)
The co-ordinates of bc = BC'^CD', cd = CD'^DA', da = DA'^AB' may be written down from Equation (2.13) by cyclic change of a, b, c, d. 3. The circles S', S'' and the coaxal system S, S', S'' and the common chord XY The circle S' passing through the four points ad, ba, cb, dc has equation (a – c)(b – d)(x2 + y2 – 1) + 4(ac – bd)x + 2(abc – abd + acd – bcd – a + b – c + d)y = 0. (3.1). Its centre O' has co-ordinates ((2(bd – ac))/((a – c)(b – d)), – (abc – abd + acd – bcd – a + b – c + d)/((a – c)(b – d))).
(3.2)
The circle S'' passing through the four points ab, bc, cd, da has equation (a – c)(d – b)(x2 + y2 – 1) + 4(ac – bd)x + 2(abc – abd + acd – bcd – a + b – c + d)y = 0. (3.3) Its centre O'' has co-ordinates ((2(bd – ac))/((a – c)(d – b ), – (abc – abd + acd – bcd – a + b – c + d)/((a – c)(d – b))).
(3.4)
The three circles S, S', S'' have a common chord XY with equation (abc + abd – acd – bcd – a – b + c + d)y = 2(cd – ab)x.
(3.5)
The actual x-co-ordinates of X and Y are very complicated and not worth recording. The line O'OO'' is perpendicular to XY through the origin and it is immediate from their coordinates given in Equations (3.2) and (3.4) that O'O = OO''.
3
4. The points U and V and the other two circles The point U = BA'^CD' and we now show U also lies on XY. The equation of BA' is (b + a)x + (ba – 1)y + b – a = 0.
(4.1)
(c + d)x + (cd – 1)y + c – d = 0.
(4.2)
The equation of CD' is
These meet at the point U with co-ordinates (x, y), where x = – (abc – abd + acd – bcd – a + b – c + d)/(abc + abd – acd – bcd + a + b – c – d), y = 2(ac – bd)/(abc + abd – acd – bcd + a + b – c – d).
(4.3) (4.4)
It may now be checked that U lies on the line XY with Equation (3.5). Similarly the point V = AB'^DC' lies on the line XY. It may also be checked that the points ab, ad, da, dc lie on a circle with equation (a – c)(d – b)(x2 + y2) + 4(ad – bc)x – 2(abc – abd – acd + bcd + a – b – c + d)y + (ab + cd + 3(ad + bc) – 4(ac + bd)) = 0.
(4.5)
Similarly the points ba, cb, bc, cd lie on a circle. The cyclic quadrilaterals generated by the four points generating the four circles are similar to each other, the proof of which is left to the reader. 5. The points F, F', F'' and their collinearity The point F is one of the exterior diagonal points of the cyclic quadrilateral ABCD. To be precise it is AB^CD. Its co-ordinates are (x, y) where x = – (abc + abd – acd – bcd – a – b + c + d)/(abc + abd – acd – bcd + a + b – c – d), (5.1) y = 2(ab – cd)/(abc + abd – acd – bcd + a + b – c – d). (5.2) The point F' is the corresponding point in S', namely ba ad^ dc cb, which from previous work is CB'^AD'. Its co-ordinates are x = – (abc + abd – acd – bcd – a – b + c + d)/(abc – abd – acd + bcd – a + b + c – d), (5.3) y = 2(ab – cd)/(abc – abd – acd + bcd – a + b + c – d). (5.4)
4
The point F'' is the corresponding point in S', namely ab bc^cd da, which again from previous work is BC'^DA'. Its co-ordinates are x = (abc + abd – acd – bcd – a – b + c + d)/(abc – abd – acd + bcd – a + b + c – d), (5.5) y = 2(cd – ab)/ abc – abd – acd + bcd – a + b + c – d). (5.6) It may now be shown that F, F', F'' lie on the line through O with equation y = (2(cd – ab))x/(abc – abd + acd – bcd – a +b – c + d). Similarly the second external diagonal points G, G', G'' are collinear.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
(5.7)
Article: CJB/2010/107
Conics through three points with Centres on a Fixed Line Christopher Bradley Abstract: Conics through three non-collinear points A, B, C having their centres on a fixed line have three radical axes, BC, CA, AB. If one takes a fixed point P on one of these axes, say BC, then the polars of P with respect to all the conics pass through another fixed point P', which as one would expect, is the harmonic conjugate of P with respect to B and C.
P
P1 S1
S4 P2
S3 Q3
A
P3 C3
P'
C4
B4 B3
C2
B2
C1
P4
B1
C0 B0
0
1
Q2
4 2
3
C
B
Q1 A4
A3 A1 A0
S0
S2
A2 To Q4
Fig. 1 Five conics through fixed points A, B, C with centres 0, 1, 2, 3, 4 lying on a fixed line 1. Introduction
1
In Fig. 1 we see five conics S0, S1, S2, S3, S4 which are representatives of the set of conics passing through three fixed points at the vertices of a triangle ABC and having their centres 0, 1, 2, 3, 4 lying on a fixed line. The fact that one of these conics is a circle and has centre at the circumcentre of ABC is incidental. We deal with the general case in the analysis that follows. Each conic Sk passes through the points A, B, C and through three other points Ak, Bk, Ck (k = 0, 1, 2, 3, 4) which are the rotations of A, B, C about k by 180 o. In the figure one may observe the polars of a fixed point P on BC with respect to each of the conics Sk all of which pass through the point P', which is the harmonic conjugate of P with respect to B and C. This result is generally true for all conics in the pencil wherever the centre of the conic lies on the fixed line and is true for not only points on the axis BC but for CA and AB also. There are thus, in effect, three radical axes. This property will be recognized by readers to be the main non-metric property of the radical axis for a system of coaxal circles. It may be seen as trivial that the points Ak, Bk, Ck form a triangle congruent to ABC and the positions of these vertices are on lines parallel to the line 01234 and such that AkAj = 2kj etc. In what follows we prove the main result stated above using areal co-ordinates with triangle of reference ABC. At the time of writing we cannot see whether a complementary pencil of conics exists as in a coaxal system of circles. 2. The line of centres and the six points on each conic We take the line of centres to have equation mnx + nly + lmz = 0.
(2.1)
A general point ‘t’ on the line has normalized co-ordinates (x, y, z), where (x, y, z) = (– l(1 + t), mt, n)/(– l(1 + t) + mt + n).
(2.2)
The point At has co-ordinates (2x – 1, 2y, 2z), where x, y, z are as in Equation (2.2). The point Bt has co-ordinates (2x, 2y – 1, 2z) and the point Ct has co-ordinates (2x, 2y, 2z – 1). 3. The conic St and the polar of P(0, s, (1 – s)) The conic St passes through A, B, C and the points At, Bt, Ct and has equation l(t + 1)(l(t + 1) + mt + n)yz + mt(l(t + 1) + mt – n)zx + n(l(t + 1) – mt + n)xy = 0.
(3.1)
We now take the fixed point P on BC to have co-ordinates (0, s, (1 – s)). The polar of P with respect to the conic St has equation (l(t + 1)(mt(s – 1) – ns) + (mt – n)(mt(s – 1) + ns))x + l(t + 1)(l(t + 1) + mt + n)((s – 1)y – sz) = 0. (3.2) It will now be observed that all such lines, whatever the value of t, pass through the point P' with co-ordinates (0, s, – (1 – s)), which is the harmonic conjugate of P with respect to B and C. 2
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
3
Article: CJB/2010/108
A Property of 3 Circles passing through a fixed Point Christopher Bradley Abstract: Let ABC be a triangle and P a fixed point not on the sides. Circle BPC meets AB at W and AC at M. Circle CPA meet BC at U and BA at N. Circle APB meets CA at V and CB at L. The perpendicular bisectors of LU, MV, NW meet at a point Q.
V
C'
A
N Q B' P
W R
L
B
U
C M
A'
Fig. 1 Circles BPC, CPA, APB
1
1. Introduction If ABC is a triangle and P is a fixed point not on any of the sides then circles BPC, CPA, APB may be drawn. Let circle BPC meet AB at W and AC at M. Let circle CPA meet BC at U and BA at N. Let circle APB meet CA at V and CB at L. Then the perpendicular bisectors of LU, MV, and NW meet at a point Q. We prove this using areal co-ordinates with ABC as triangle reference. An interesting corollary is that, if P is the centroid G, then the point Q is the circumcentre O. 2. The three circles Let P have areal co-ordinates (l, m, n). B and C have co-ordinates (0, 1, 0) and (0, 0, 1). The equation of the circle BPC may now be calculated and is (a2mn + b2nl + c2lm)x2 – a2l(l + m + n)yz + (a2mn – b2l(l + m) + c2lm)zx + (a2mn + b2nl – c2l(n + l))xy = 0. (2.1) Circles CPA and APB have equations that may be written down from Equation (2.1) by cyclic change of x, y, z and a, b, c and l, m, n. 3. The points L, M, N and U, V, W on the sides Circle BPC with Equation (2.1) meets CA, y = 0, at M and AB, z = 0 at W. The co-ordinates of M are (b2l(l + m) – c2lm – a2mn, 0, a2mn + b2nl + c2lm).
(3.1)
The co-ordinates of W are (– a2mn – b2nl + c2l(n + l), a2mn + b2nl + c2lm, 0).
(3.2)
Points N, L and U, V have co-ordinates that may be written down from those in (3.1) and (3.2) by cyclic change of x, y, z and a, b, c and l, m, n. 4. The perpendicular bisector of NW The equation of the line CH is (c2 + a2 – b2)y = (b2 + c2 – a2)x.
(4.1)
The line parallel to this through the midpoint of NW is of the form s(x + y + z) + (c2 + a2 – b2)y – (b2 + c2 – a2)x = 0,
(4.2)
where s is a constant to be determined from inserting the co-ordinates of the midpoint of NW. 2
The result is (a2mn(l – m) + b2nl(l – m) – c2lm(2m + n))x + (a2mn(l – m) + b2nl(l – m) + c2lm(2l + n))y + (b2l(lm + ln + m2) + c2lm(l – m) – a2lm(l2 + lm + mn)z = 0. (4.3) The perpendicular bisectors of LU and MV may be obtained from Equation (4.3) by cyclic change of x, y, z and a, b, c and l, m, n. 5. The point of concurrence of the three perpendicular bisectors It may now be shown that the three perpendicular bisectors meet at the point Q with co-ordinates (x, y, z), where x = mn(l2 + 2mn)a4 + l2n(m – n)b4 – l2m(m – n)c4 + l2(m – n)2b2c2 – l2m(m + 2n)c2a2 – l2n(2m + n)a2b2. (5.1) 2 4 2 4 2 4 2 2 2 2 2 2 2 y = m n(l – n)a + nl(2nl + m )b + m l(n – l)c – m l(l + 2n)b c + m (l – n) )c a – m2n (2l + n)a2b2. (5.2) 2 4 2 4 2 4 2 2 2 2 2 2 z = mn (l – m)a – ln (l – m)b + lm(2lm + n )c – ln (l + 2m)b c – mn (2l + m)c a + n2(l – m)2a2b2. (5.3) When l = m = n = 1/3 and P lies at the centroid G then it may be checked that Q lies at the circumcentre O. Flat 4, Terrill Court 12-14, Apsley Road, BRISTOL. BS8 3SP.
3
Article: CJB/109/2010
Four Perspectives given a Triangle and its Circumcircle Christopher Bradley Abstract: Given a triangle, its circumcircle and a point not on the circumcircle, it is shown how by varying the position of the point along a line four perspectives may be created. V
To V A
C
B
M
K F
E
N
T
S
D Q
R To J To U P
Fig. 1 U Triangle ABC and a selected point P produce four perspectives 1. Introduction Given triangle ABC and its circumcircle S a point P not on the circumcircle may be adjusted so that AKP is a straight line, where K = BF^CE and E = BP^S and F = CP^S. We show in subsequent sections, using Cartesian co-ordinates, that the condition for this to happen is that P should lie on a line determined by the positions of A, B and C. When P is chosen on this line 1
then four perspectives are created (i) ABC and KEF with vertex P, (ii) ABC and PFE with vertex K, (iii) ABC and DEF with vertex P and (iv) ABC and DFE with vertex K. The four axes of perspective have a common point T = EF^BC and K is the harmonic conjugate of P with respect to A and D = AP^S. 2. Lines AP, BP, CP and points D, E, F Let P have co-ordinates (d, e) with d2 + e2 ≠ 1. Later it is shown that for K to be on AP it is necessary that P should lie on a line determined by A, B, C only. The co-ordinates of points A, B, C are A((1 – a2)/(1 + a2), 2a/(1 + a2)), with B, C similar but with parameters b, c rather than a. The equation of the line AP may now be obtained and is ((1 + a2)e – 2a)x – ((1 + a2)d – (1 – a2))y – (1 – a2)e + 2ad = 0.
(2.1)
The equations of BP, CP may be obtained from Equation (2.1) by replacing a with b and c respectively. The point D where AP meets the circumcircle S has co-ordinates (x, y), where x = (a2( (d + 1)2 – e2) – 4ade + e2 – (d – 1)2)/(a2( (d + 1)2 + e2) – 4be + (d – 1)2 + e2), (2.2) 2 2 2 2 2 2 2 2 y = (2(a (1 + d)e + a((d – 1) – e ) – e(d – 1)))/(a ( (d + 1) + e ) – 4be + (d – 1) + e ). (2.3) Points E and F have similar co-ordinates but with b, c respectively replacing a. 3. Lines CE, BF and the point K The equations of the lines CE and BF may now be obtained and are CE: (b(ce – d – 1) + c(d – 1) + e)x – (b(c(d + 1) + e) – ce + d – 1)y + b(ce + d + 1) + c(d – 1) – e = 0.
(3.1)
(b(ce + d – 1) – c(d + 1) + e)x – (b(c(d + 1) – e) + ce + d – 1)y + b(ce + d – 1) + c(d + 1) – e = 0.
(3.2)
BF:
These lines meet at the point K with co-ordinates (x, y), where x = (b(c(d – e2 + 1) – de) – cde + d + e2 – 1)/(b(c(d2 + d + e2) – e) – ce + d2 – d + e2), (3.3) 2 2 2 y = (b(d + 1)(ce + d – 1) + c(d + 1)(d – 1) – e(d – 1))/(b(c(d + d + e ) – e) – ce + d – d + e2). (3.4) 4. The condition that K lies on AP Substitution of the co-ordinates of K in Equations (3.3) and (3.4) into the equation of the line AP in Equation (2.1) provides the condition that K lies on AP and this surprisingly is a linear relation
2
between d and e, showing that the locus of the position of P that implies K lies on AP is a line with equation 2(a2 – bc)e – (a2(b + c) – 2a(1 + bc) + b + c)d = (a2(b + c) + 2a(1 – bc) – b – c). (4.1) David Monk (private communication) has pointed out that this line is the line through A passing through the symmedian point. 5. The four perspectives An immediate consequence of K lying on AP is that K and P are then harmonic conjugates of A and D. This is because (AKDP) = C(AEDF) = B(AEDF) = (APDK). It is also clear that when K lies on ADP then four perspectives are created. These are (i) Triangles ABC and KEF with vertex P and axis TSR; (ii) Triangles ABC and PFE with vertex K and axis TUV; (iii) Triangles ABC and DEF with vertex P and axis TMN; (iv) Triangles ABC and DFE with vertex K and axis TQJ. See Fig. 1 where the positions of points T, S, U, V, M, N, Q, J lie. Note that T lies on all four axes. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2010/110
Three Concurrent Pascal lines and an Auxiliary Conic Christopher Bradley
W M
L 2' A
E
To J
F 3'
P V C
B
To Y
To 2
D
U 1 3
1' To X
To Z
Y
Fig. 1 Three Pascal lines and an Auxiliary Conic
1. Introduction X
A hexagon inscribed in a conic is created by a triangle ABC and the triangle inscribed in the circumscribing conic S of ABC by a second triangle DEF which is in perspective with ABC through a vertex P. The points of intersection of the nine lines AD, AE, AF, BD, BE, BF, CD, CE, CF are created and these are sixteen in number (or eighteen if you count the appearance of P 1
2
three times). The points 1 = BE^CD, 2 = CF^AE, 3 = AD^BF lie on a Pascal line of the hexagon. The points L = CF^BD, M = AD^CE and N = BE^AF lie on a second Pascal line of the hexagon and the points X = BF^CE, Y = CD^AF and Z =, AE^BE lie on a third Pascal line of the hexagon. The three Pascal lines are concurrent at the Steiner point J. The points 1' = BD^CE, 2' = CE^AF, 3' = AF^BD, U = BF^CD, V = CD^AE, W = AE^BF turn out to lie on a conic Σ. It should be appreciated that such an auxiliary conic does not in general appear in a Pascal configuration and Cabri indicates that it does so only when the perspective exists. An additional bonus is that the line XYZ is also the polar of P with respect to both conics S and Σ. The configuration is analysed using homogeneous projective co-ordinates with ABC as triangle of reference and P having co-ordinates (1, 1, 1). 2. The points D, E, F and the lines AD, BE, CF, AE, AF, BD, BF, CD, CE We take the conic S to have equation fyz + gzx + hxy = 0
(2.1)
terms in x2, y2, z2 being absent since S passes through A, B, C. We take P to be the unit point (1, 1, 1). Lines AP, BP, CP therefore have equations y = z, z = x, x = y respectively. AP meets S at the point D(– f, g + h, g + h) and E, F have co-ordinates E(h + f, – g, h + f), F(f + g, f + g, – h) respectively. Lines AE, AF, BD, BF, CD, CE have equations as follows: AE: (h + f)y + gz = 0, AF: hy + (f + g)z = 0, BD: fz + (g + h)x = 0, BF: (f + g)z + hx = 0, CD: (g + h)x + fy = 0, CE: gx + (h + f)y = 0.
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7)
3. The points 1, 2, 3 and the Pascal line 123 Point 1 = CD^BE has co-ordinates (f, – (g + h), f), point 2 = AE^CF has co-ordinates (g, g, – (h + f)) and point 3 = BF^AD has co-ordinates (– (f + g), h, h). These points are collinear on the Pascal line 123 with equation hx + fy + gz = 0. 4. Points L, M, N and the Pascal line LMN Point L = CF^BD has co-ordinates (f, f, – (g + h)), point M = AD^CE has co-ordinates 2
(3.1)
(– (h + f), g, g) and point N has co-ordinates (h, – (f + g), h). These points are collinear on the Pascal line LMN with equation gx + hy + fz = 0.
(4.1)
5. Points X, Y, Z| and the Pascal line XYZ and the point J Point X = BF^CE has co-ordinates (– (f + g)(f + h), g(f + g), h(f + h)), point Y = CD^AF has coordinates (f(g + f), – (g + h)(g + f), h(g + h)) and point Z = AE^BD has co-ordinates (f(h + f), g(h + g), – (h + f)(h + g)). These points are collinear on the Pascal line XYZ with equation (g + h)x + (h + f)y + (f + g)z = 0.
(5.1)
It is obvious from Equations (3.1), (4.1), (5.1) that the Pascal lines 123, LMN, XYZ are concurrent. Their point of concurrence is the Steiner point J, which has co-ordinates J(f2 – gh, g2 – hf, h2 – fg). It may easily be checked that XYZ is also the polar of P with respect to the conic S. 6. The points 1', 2', 3', U, V, W and the conic Σ containing these points Point 1' = BD^CE has co-ordinates (– f(f + h), fg, (f + h)(g + h)), point 2' = CE^AF has coordinates ((g + f)(h + f), – g(g + f), gh) and point 3' = AF^BD has co-ordinates (hf, (h + g)(f + g), – h(h + g)). Point U = BF^CD has co-ordinates (– f(f + g), (f + g)(g + h), hf), point V = CD^AE has coordinates (fg, – g(g + h), (g + h)(f + g)) and point W = AE^BF has co-ordinates ((h + f)(f + g), gh, – h(h + f)). These six points lie on the conic Σ with equation ux2 + vy2 + wz2 + 2pyz + 2qzx + 2rxy = 0
(6.1)
where u = 2fgh(g + h), v = 2fgh(h + f), w = 2fgh(f + g), p = f3(g + h) + fgh(f + g + h) + f2(g2 + h2), q = g3(h + f) + fgh(f + g + h) + g2(h2 + f2), r = h3(f + g) + fgh (f + g + h) + h2(f2 + g2). 3
(6.2) (6.3) (6.4) (6.5) (6.6) (6.7)
It may now be shown that the line XYZ is also the polar of P with respect to the conic Σ.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
CJB: Article/2010/111
Perpendicular bisectors of Three Radii of a Circle Christopher Bradley
N3
A L3 N2 L2 O M3
B
M1
N1
L1
C
M2
Fig. 1 Transversal formed by the perpendicular bisectors of OA, OB, OC 1
1. Introduction We suppose the three radii are labelled OA, OB, OC so that the circle in question is the circumcircle of ABC. Now let the perpendicular bisectors of OA, OB, OC meet the lines BC, CA, AB respectively at L1, M2, N3. We establish that these points are collinear. This means that the six other points of intersection of the perpendicular bisectors with the sides of the triangle ABC lie on a conic. Cabri indicates that O and H are the only points internal or external to triangle ABC for which the construction leads to a collinearity (and H is singular, as the line then is the line at infinity). We use Cartesian co-ordinates with circle ABC as the unit circle and O the origin. 2. The Analysis We take A to have co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)). B and C have similar expressions for their co-ordinates, but with parameters b, c replacing a. The line OA has equation 2ax = (1 – a2)y. The perpendicular bisector of OA has equation 2(a2 – 1)x – 4ay + (1 + a2) = 0.
(2.1)
(2.2)
The equation of BC is (1 – bc)x + (b + c)y = (1 + bc). The intersection of these last two points is the point L1 with co-ordinates (x, y), where x = (4a(1 + bc) – (1 + a2)(b + c))/(2(2a(1 – bc) + (a2 – 1)(b + c)), y = ((a2 – 3)bc + 3a2 – 1)/(2(2a(1 – bc) + (a2 – 1)(b + c)).
(2.3)
(2.4) (2.5)
The points M2 and N3 have co-ordinates that may be obtained from Equations (2.4) and (2.5) by cyclic change of a, b, c. The three points L1, M2, N3 may now be shown to lie on the line with equation lx + my + n = 0, where
2
(2.6)
l = 2(3a2b2c2 – 3(b2c2 + c2a2 + a2b2) + 4abc(a + b + c) – 4(bc + ca + ab) + 3(a2 + b2 + c2) – 3), m = – 4((abc(a + b + c) – (a2b + a2c + b2c + b2a + c2a + c2b) + 12abc – (a + b + c)) , n = 5a2b2c2 – 3(b2c2 + c2a2 + a2b2) + 8abc(a + b + c) + 8(bc + ca + ab) – 3(a2 + b2 + c2) + 5. When 3 lines intersect the three sides of a triangle there are 9 points of intersection, which, in general, lie on a cubic. However, if three lie on a line, it is a general result that the other six must lie on a curve of degree two (or two more lines), that is a conic. See Fig.1 for the conic L2L3M3M1N1N2.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2010/112
Parallels from the feet of Cevians create a Conic Christopher Bradley Abstract: From the feet of Cevians through a point parallels to the other two sides are drawn. The six points created on the sides of the triangle lie on a conic.
A
L3 N2 N M3
M P
L2
M1 B
N1
L
C
Fig. 1 The conic formed by parallels from the feet of Cevians 1. Introduction Let ABC be a triangle and P a point not on the sides of ABC. Cevians AP, BP, CP are drawn to meet the sides BC, CA, AB respectively at L, M, N. Lines through L are drawn parallel to AB and AC to meet the sides AC and AB respectively at L2 and L3. Points M1 on BC and M3 on AB 1
are similarly defined as are points N1 on BC and N2 on CA. We establish the result that these six points lie on a conic. 2. The six points If P has co-ordinates (l, m, n), l, m, n ≠ 0 the feet of the Cevians have co-ordinates L(0, m, n), M(l, 0, n), N(l, m, 0). A line parallel to AB has equation of the form z + k(x + y + z) = 0, where k depends on the point through which the parallel is drawn. If the parallel passes through L(0, m, n) then the appropriate value of k = – n/(m + n) and the line therefore has equation n(x + y) – mz = 0. (2.1) This line meets AC with equation y = 0 at the point L2 with co-ordinates (m, 0, n). Similarly L3 has co-ordinates (m, n, 0). We can now write down the co-ordinates of the other four points by cyclic change of x, y, z and l, m, n. They are M3(l, n, 0), M1(0, n, l), N1(0, m, l) and N2(m, 0, l). 3. The conic through the six points The equation of the conic is ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where u = 2ln2, v = 2ml2, w = 2nm2, f = – lm(m + n), g = – mn(n + l), h = – nl(l + m).
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
The conic is a circle if and only if the sides a, b, c are related to l, m, n by the equations n(a2 – n2l) = l(b2 – l2m) = m(c2 – m2n).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
2
(3.8)
Article: CJB/2010/113
Circles passing through the Miquel Point of the feet of Cevians Christopher Bradley
A W
X N
T S Q
O P R
V
M
Z
Y L
B
C
U
Fig. 1 Many circles passing through a Miquel point Abstract: If LMN are the feet of a set of Cevians, then circles AMN, BNL, CLM meet at a Miquel point Q. If now circles BQC, CQA, AQB are drawn and circles AMN and BQC meet at R with S, T similarly defined, then circle RST passes through the Miquel point Q. The centres of the seven circles also exhibit some remarkable properties showing Q to be the Miquel point of a second triangle, a cascade process that can be carried on indefinitely. 1
1. Introduction A point P is chosen, which may be inside or outside triangle ABC, and the Cevians AP, BP, CP are drawn meeting BC, CA, AB respectively in the points L, M, N. The Miquel circles AMN, BNL, CLM meet at a point Q, known as the Miquel point of L, M, N. Circles BQC, CQA, AQB may now be drawn. The second point of intersection (other than Q) of circles AMN and BQC is denoted by R and those of circles BNL and CQA and of CLM and AQB are denoted respectively by S and T. We prove that R, S, T lie on a circle, centre O, that passes through Q. Cabri indicates that this happens only when L, M, N are the feet of Cevians. Furthermore AR, BS, CT are concurrent at a point J that lies on circle RST. Circles AMN, BNL, CLM have centres X, Y, Z|. Circles BQC, CQA, AQB have centres U, V, W. We prove UX, VY, WZ pass through O and that points X, Y, Z lie on the lines VW, WU, UV respectively. Thus X, Y, Z are the feet of Cevians of triangle UVW. Cabri indicates that the Miquel point of X, Y, Z relative to triangle UVW is again the point Q. This means there is a cascade process in which the construction may be repeated indefinitely with an ever increasing number of circles all passing through Q. When P is at the orthocentre H, then Q lies at H also. When P lies at the centroid G, then Q lies at the circumcentre of ABC. In what follows we use arealk co-ordinates with ABC as triangle of reference. 2. Circles AMN, BNL, CLM and the Miquel point Q If P has co-ordinates P(l, m, n) then the feet of the Cevians AP, BP, CP have co-ordinates respectively L(0, m, n), M(l, 0, n), N(l, m, 0). The equation of the circle AMN is a2yz + b2zx + c2xy + (px + qy + rz)(x + y + z) = 0,
(2.1)
where p = 0, q = – c2l/(l + m), r = – b2l/(l + n).
(2.2)
The equations of circles BNL, CLM now follow from Equation (2.2) by cyclic change of p, q, r and l, m, n and a, b, c. Point Q may now be calculated as the common point of intersection of the above three circles and its x-co-ordinate is x = – a4mn (l + m)2(l + n)2 + a2b2nl(l + n)(n + m)(l + m)2 + a2c2ml((l + m)(m + n)(l + n)2. (2.3) 2
The y- and z- co-ordinates of Q follow by cyclic change of a, b, c amd l, m, n. 3. Circles BCQ, CAQ, ABQ and the points R, S, T We may now obtain the equations of circles BCQ, CAQ, ABQ. Circle BCQ has equation (2.1) with q = 0, r = 0 and p = 2b2c2/ (a2mn(l + m)(l + n) – l(m + n)(b2n(l + m) + c2m(l + n))). (3.1) The equations of circles CAQ and ABQ may be obtained by cyclic change of p, q, r and a, b, c and l, m, n. The intersection of circles AMN and BCQ is the point R and its co-ordinates (x, y, z) turn out to be x = l(m + n)(b2n(l + m) + c2m(l + n) – a2mn(l + m)(l + n), (3.2) 2 2 y = 2b mn (l + m). (3.3) 2 2 z = 2c m n(l + n). (3.4) The co-ordinates of S and T may be obtained from Equations (3.2) to (3.4) by cyclic changes of x, y, z and l, m, n and a, b, c. The equation of the circle RST may now be obtained and has Equation (2.1) with p = (1/k)(2b2c2lmn(m + n)), (3.5) 2 2 q = (1/k)(2c a lmn(n + l)), (3.6) r = (1/k)(2a2b2lmn(l + m)), (3.7) where k = a2mn(l + m)(l + n) + b2nl(m + n)(m + l) + c2lm(n + l)(n + m). (3.8) It may now be shown that the Miquel point Q with co-ordinates given by Equation (2.3) lies on the circle RST. Cabri indicates that this happens only when the points L, M, N are the feet of Cevians. The lines AR, BS, CT are concurrent at a point J with co-ordinates (x, y, z), where x = a2/(l(m + n)), y = b2/(m(n + l)), z = c2/(n(l + m)). It may now be shown that J lies on circle RST with Equations (3.5) – (3.8). 4. The centres of circles RST, AMN, BNL, CLM, BCQ, CAQ, ABQ The circle with Equation (2.1) has co-ordinates (x, y, z), where 3
(3.9)
x = a4 – a2(b2 + c2 + 2p – q – r) + (b2 – c2)(q – r), y = b4 – b2(c2 + a2 + 2q – r – p) + (c2 – a2)(r – p), z = c4 – c2(a2 + b2 + 2r – p – q) + (a2 – b2(p – q).
(4.1) (4.2) (4.3)
Equations (4.1) – (4.3) enable us to determine the centre O of circle RST, the centres X, Y, Z of the circles AMN, BNL, CLM respectively and the centre U, V, W of circles BCQ, CAQ, ABQ. It may now be checked that lines XU, YV, ZW are concurrent at O, that points V, X, W, are collinear as are W, Y, U, and U, Z, V. It follows that UVW is a triangle in which X, Y, Z are the feet of Cevians of the point O. And so there exists a Miquel point, which is the intersection of circles WXY, UYZ, VZX. Interestingly, this point turns out to be Q. We do not record details of the calculation to verify this, as they are exceedingly complicated technically. The construction involving A, B, C, L, M, N, P, Q may now be repeated with points U, V, W, X, Y, Z, O, Q and indeed may be repeated indefinitely and with each such construction there are seven circles all passing through Q. When P lies at the orthocentre H, then Q lies at H also. When P lies at the centroid G, then Q lies at the circumcentre of ABC.
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Article: CJB/2010/114
Properties of a Particular Tucker Circle Christopher Bradley Abstract: Three circles determine a particular Tucker circle and their centres exhibit some interesting properties.
N3 D
M3
N2 To M2
A W N1
V K
B
E
C L3
F M1
U L1
Fig. 1 A Tucker circle and its centres
1
L2
1. Introduction Given a triangle ABC and its symmedian point K a construction of Fig. 1 starts by drawing the circles BKC, CKA, AKB. Circle BKC meets AB at N1 and AC at M1. Points L2 and N2 on CA are similarly defined by circle CKA and points M3 and L3 are similarly defined by circle AKB. The lines L2M1 and N1L3 meet at L1, lines M3N2 and L2N1 meet at M2 and ;ines N1L3 and M3N2 meet at N3. We prove that triangle L1M2N3 is an enlargement by a factor of two and a rotation of 180o of triangle ABC about K. The centres of circles BKC, CKA, AKB are denoted by U, V, W and the centres of circles AM3N2 , BN1L3, CL2M1 are denoted by D, E, F. We prove that a conic passes through the six points U, V, W, D, E, F. We use areal co-ordinates throughout with ABC as triangle of reference. 2. Circles BKC, CKA, AKB and their intersections with the sides of ABC All circles have an equation of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(2.1)
where u, v, w are determined by the three points defining the circle. The circle BKC has u = – 3b2c2/(a2 + b2 + c2), v = 0, w = 0 and hence has an equation 3b2c2x2 – a2(a2 + b2 + c2)yz – b2(a2 + b2 – 2c2)zx – c2(c2 + a2 – 2b2)xy = 0. (2.2) It meets CA at M1 with co-ordinates (a2 + b2 – 2c2, 0, 3c2) and it meets AB at N1 with coordinates (c2 + a2 – 2b2, 3b2, 0). The equations of circles CKA and AKB have equations that may be determined from Equation (2.2) by cyclic change of x, y, z and a, b, c. Point N2 has co-ordinates (3a2, b2 + c2 – 2a2, 0) and point L2 has co-ordinates (0, a2 + b2 – 2c2, 3c2). Point L3 has coordinates (0, 3b2, c2 + a2 – 2b2) and point M3 has co-ordinaters (3a2, 0, b2 + c2 – 2a2). 3. The Tucker circle and the triangle L1M2N3 The Tucker circle passes through the six points defined in Section 2 and has an Equation of the form (2.1) with u = 3b2c2(2a2 – b2 – c2)/(a2 + b2 + c2)2, (3.1) v = 3c2a2(2b2 – c2 – a2)/(a2 + b2 + c2)2, (3.2) 2 2 2 2 2 2 2 2 w = 3a b (2c – a – b )/(a + b + c ). (3.3) The equation of the line M3N2 is (2a2 – b2 – c2)x + 3a2(y + z) = 0.
2
(3.4)
This line is parallel to BC and similarly N1L3 is parallel to CA and L2M1 ias parallel to AB, verifying that this circle is indeed a Tucker circle. The equations of N1L3 is and the equation of L2M1 is
(2b2 – c2 – a2)y + 3b2(z + x) = 0
(3.5)
(2c2 – a2 – b2)z + 3c2(x + y) = 0.
(3.6)
These last two lines meet at the point L1 with co-ordinates (a2 – 2b2 – 2c2, 3b2, 3c2). Similarly M2 has co-ordinates (3a2, b2 – 2c2 – 2a2, 3c2) and N3 has co-ordinates (3a2, 3b2, c2 – 2a2 – 2b2). 4. Circles AM3N2, BN1L3, CL2M1 and the circle centres U, V, W, D, E, F The circle AM3N2 has equation in the form (2.1) with u = 0, v = – 3c2a2/(a2 + b2 + c2), w = – 3a2b2/(a2 + b2 + c2).
(4.1) (4.2) (4.3)
The equations of circles BN1L3 and CL2 M1 may be obtained from that of circle AM3N2 by cyclic change of x, y, z and u, v, w and a, b, c. The co-ordinates of the centre of a circle with equation of the form (2.1) are given by x = a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w), (4.4) 4 2 2 2 2 2 y = b – b (c + a + 2v – w – u) + (c – a )(w – u), (4.5) z = c4 – c2(a2 + b2 + 2w – u – v) + (a2 – b2)(u – v). (4.6) Since we have the values of u, v, w for each of the six circles we are in a position to write down the co-ordinates of U, V, W, D, E, F. For U these are x = a2(a4- b4 + 4b2c2 – c4), (4.7) y = – b2(a4 + 5a2c2 – b4 + 3b2c2 – 2c4), (4.8) 2 4 2 2 4 2 2 4 z = – c (a + 5a b – 2b + 3b c – c ). (4.9) The co-ordinates of V and W may be written down from Equations (4.7) – (4.9) by cyclic change of x, y, z abd a, b, c. The co-ordinates of D are x = a2(a4 – 3a2(b2 + c2) + 2(b4 – 4b2c2 + c4)), y = b2(a2 – b2 + c2)(2a2 – b2 – c2), z = c2(a2 + b2 – c2)(2a2 – b2 – c2).
(4.10) (4.11) (4.12)
The co-ordinates of E and F may be written down from Equations (4.10) – (4.12) by cyclic change of x, y, z and a, b, c. 5. The conic through the six circle centres U, V, W, D, E, F 3
This has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(5.1)
where u = – 2b2c2(2a2 – b2 – c2)(a4(b2 + c2) + 2a2(b4 + 3b2c2 + c4) + b6 – b4c2 – b2c4 + c6) (5.2) and v, w follow by cyclic change of a, b, c. Also f = – a2b2c2(12a6 + 7a4(b2 + c2) – 4a2(b4 – 7b2c2 + c4) + b6 + 3b4c2 + 3b2c4 + c6)
(5.3)
and g, h follow by cyclic change of a, b, c. 6. The relationship between triangles ABC and L1M2N3 We find the displacement AK to be (1/(a2 + b2 + c2))(– (b2 + c2), b2, c2) and KL1 = 2AK. Similarly KM2 = 2BK and KN3 = 2CK. Thus triangle L1M2N3 is an enlargement by a factor of two and a rotation of 180o of triangle ABC about K. We do not supply the details but it may be shown that triangle DEF and UVW are congruent and that they are a 180o rotation of each other about a point J, where KJ = (1/4)KO, O being the circumcentre of ABC. Thus J lies on the Brocard axis (at a point that I confess I have not encountered in any of my previous configurations. See Fig. 2.
4
N3 D
M3
N2 To M2
A W N1
V K J
B
O E
C L3
F M1
U L1
Fig. 2 The point J on the Brocard axis
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L2
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Article: CJB/115/2010
Six Circles and their Centres Christopher Bradley Abstract: In triangle ABC centroid G, circles BGC, CGA, AGB are drawn meeting the sides in six points that lie on a conic. Three more circles are drawn and the six circle centres exhibit some interesting properties.
M3
D W
A N2
V O N1
E B
G
L2
L3
C
F U
Fig. 1 Circle centres forming a pair of congruent triangles 1
M1
1. Introduction Given a triangle ABC and its centroid G a construction of Fig. 1 starts by drawing the circles BGC, CGA, AGB. Circle BGC meets AB at N1 and AC at M1. Points L2 and N2 on CA are similarly defined by circle CGA and points M3 and L3 are similarly defined by circle AGB. The centres of circles BGC, CGA, AGB are denoted by U, V, W and the centres of circles AM3N2 , BN1L3, CL2M1 are denoted by D, E, F. We prove that a conic passes through the six points U, V, W, D, E, F. We also prove that DU, EV, FW are concurrent at the circumcentre of triangle ABC,that triangles DEF and UVW are congruent and that triangles DBC, ECA, FAB are isosceles. Areal co-ordinates are used throughout with ABC as triangle of reference. 2. Circles BGC, CGA, AGB and their intersections with the sides of ABC Circles in areal co-ordinates have equations of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(2.1)
where u, v, w are constants to be determined from the co-ordinates of the three points defining the circle. The circle BGC has v = w = 0 and u = –(1/3)(a2 + b2 + c2) and so has equation (a2 + b2 + c2)x2 – 3a2yz + (c2 + a2 – 2b2)zx + (a2 + b2 – 2c2)xy = 0. (2.2) The equations of circles CGA and AGB may be written down from Equation (2.2) by cyclic change of x, y, z and a, b, c. Circle BGC meets AB at the point N1 with co-ordinates N1(2c2 – a2 – b2, a2 + b2 + c2, 0). And circle BGC meets CA at the point M1 with co-ordinates M1(2b2 – c2 – a2, 0, a2 + b2 + c2). The co-ordinates of points L2, M3 may be obtained from those of N1 by cyclic change of x, y, z and a, b, c. Similarly those of N2 and L3 may be obtained from those of M1 by cyclic change of x, y, z and a, b, c.
3. The conic passing through N 1, L2, M3, M1, N2, L3 Conics in areal co-ordinates have equations of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(3.1)
where the constants u, v, w, f, g, h are determined up to a common factor from five of the points on the conic, it then being checked that the sixth point also lies on the conic. The conic passing through N 1, L2, M3, M1, N2, L3 has u = 2(a4 + b4 – 2c4 – b2c2 – c2a2 + 2a2b2)(a4 + c4 – 2b4 – b2c2 – a2b2 + 2a2c2) all times (2a4 + a2(b2 + c2) – (b2 + c2)2). (3.2) 2
The values of v and w may be written down from Equation (3.2) by cyclic change of a, b, c. Also f = – (a4 + b4 – 2c4 – b2c2 – c2a2 + 2a2b2)( a4 + c4 – 2b4 – b2c2 – a2b2 + 2a2c2) all times (5a4 – 2a2(b2 + c2) + 2(b2 + c2)2). (3.3) The values of g and h may be written down from Equation (3.3) by cyclic change of a, b, c. If the centroid is replaced by any point P not on the sides of the triangle ABC and circles BPC, CPA, APB intersect the sides BC, CA, AB again in six points, Cabri indicates that a conic always passes through the six points. 4. The three circles AM3N2, BN1L3, CL2M1 and the centres U, V, W, D, E, F Circle AM3N2 has an equation of the form (2.1) with u = 0, v = w = – (1/3)(a2 + b2 + c2).
(4.1)
and hence its equation is (a2 + b2 + c2)(y2 + z2) + (2b2 + 2c2 – a2)yz +(c2 + a2 – b2)zx + (a2 + b2 – 2c2)xy = 0. (4.2) Circles BN1L3 and CL2M1 have equations that may be written down from (4.2) by cyclic change of x, y, z and a, b, c. With the equation of a conic in the form (3.1) the co-ordinates of the centre are (x, y, z), where x = a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w), (4.3) and where y, z may be obtained from (4.1) by cyclic change of a, b, c and u, v, w. It soon follows that the co-ordinates of U are (x, y, z), where x = a2(5a2 – b2 – c2), y = – (a4 + 5a2b2 – 2b4 + 3b2c2 – c4), z = – (a4 + 5a2c2 – b4 + 3b2c2 – 2c2).
(4.4)
The co-ordinates of V and W follow from equations(4.4) by cyclic change of x, y, z and a, b, c. It now follows that the co-ordinates of D are x = a2(a2 – 5b2 – 5c2), y = a4 + 4b4 – c4 – a2b2 – 3b2c2, z = a4 – b4 + 4c4 – a2c2 – 3b2c2.
(4.5)
The co-ordinates of E and F follow from equations (4.5) by cyclic change of x, y, z and a, b, c. 3
5. The lines DU, EV, FW are concurrent at the circumcentre O of ABC From the co-ordinates of D, U and O it may now be shown that the displacements DO and OU are equal, the displacements being proportional to (x, y, z) where x = 2a2(a2 + b2 + c2), y = – a4 – 2a2b2 – b4 + c4, (5.1) z = – a4 – 2a2c2 + b4 – c4. It also is the case that DOU is perpendicular to BC and so it follows that DB = DC. It is also the case that EV and FW pass through O and that EC = EA and FA = FB. Also it follows that triangles UVW and DEF are congruent by a rotation of 180o about O. It should be noted that if G is altered to a general point P not on the sides of ABC then Cabri indicates that triangles DEF and UVW are congruent and that DU, EV , FW are concurrent.
6. The conic through U, V, W, D, E, F When written in the form of equation (3.1) the values of u, v, w, f, g, h are as follows: u = – 2(a4 – 8b4 – 8c4 + 2a2b2 + 2a2c2 – 11b2c2), (6.1) with v, w as in (6.1) after cyclic change of a, b, c and f = – 11a4 – 2b4 – 2c4 – 13a2b2 – 13a2c2 + 14b2c2, with g, h as in (6.2) after cycic change of a, b, c.
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(6.2)
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Article: CJB/116/2010
The Geometry of the Brocard Axis and Associated Conics Christopher Bradley
N W
X M2 F
T
L2
A A' S
F'
E V L1
O R E'
B' D'
B
Q N1
D
K C'
C
M1
L Z
N2 U
Y
M
Fig.1 The Brocard axis and its conics
Abstract: From a triangle and its symmedian point K and its circumcentre O, various triangles, the Brocard inellipse and two further conics are constructed having their centres on the Brocard axis KO. Two porisms are defined and a special line is identified. Most, if not all, of the content of this article is known and indeed should be known by all geometers. 1
1. Introduction Triangle ABC and its symmedian point K are given. Points D, E, F are the intersections of the lines AK, BK, CK with the circumcircle Σ. The tangents are drawn at A, B, C creating a triangle UVW, where U, V, W lie on AK, BK, CK respectively. Tangents are also drawn at D, E, F creating a second triangle XYZ and X, Y, Z also lie on AK, BK, CK respectively. It follows that triangles ABC, DEF are in perspective, vertex K and triangles UVW and XYZ are in perspective, vertex also K. We prove the result that X, W, Y, U, Z, V lie on a conic, centre T that lies on the Brocard axis KO. The nine intersections of triangles UVW and XYZ lie six on a second conic, whose centre S lies on the Brocard axis KO, and the remaining three on a line LMN. The points on the conic are L1, L2 on a line through L, M1, M2 on a line through M and N1, N2 on a line through N. The line LMN is the polar of K with respect to Σ and the points on it have many properties. L, for example, lies on the lines BC, VW, YZ, EF, L1L2, B'C', E'F' and the point labelled Q lies on AF, BE, CD showing that it is the vertex of perspective of triangles ABC and FED. Such a perspective is called a reverse perspective and the triangles ABC and DEF are, in fact, in triple reverse perspective, ABC being in perspective with DFE, EDF and FED. The vertices of these perspectives all lie on line LMN. Further lines passing through Q are L2N1, N2L1, XW, YV, ZU. This shows that triangles XYZ and UVW besides being in direct perspective with vertex K are also in reverse perspective with vertex Q. In fact, they too are in triple reverse perspective, with vertices the same as the reverse perspectives of ABC and DEF. The intersections of triangles ABC and DEF do not appear in Fig. 1, but six of them also lie on a conic and three of them, such as AB and DE meet at the point N on the polar line, AC and DF meet at M and BC and EF at L. Thus LMN is the Desargues’ line of perspective of the triangles ABC and DEF. Also if the lines AD, BE, CF meet BC, CA, AB at D', E', F' respectively and AD, BE, CF meet EF, FD, DE at A', B', C' respectively, then A', B', C', D', E', F' lie on a conic touching the sides of ABC and DEF. This is sometimes known as the Brocard inellipse. Its centre R lies on the Brocard axis KO. Thus one has two triangles ABC and DEF with vertices on Σ and sides touching the Brocard inellipse, and hence a porism is created, usually known as the Brocard porism. It is also the case that triangles XYZ and DEF have their vertices on the conic centre T, which by the initial construction have their sides touching Σ and hence we have a second porism, which I call the outer Brocard porism. In what follows we use areal co-ordinates with ABC as triangle of reference to establish many of the above results, some, however, being left for the reader. 2
2. The points D, E, F and U, V, W The point D is the intersection of the line AK with the circumcircle Σ. Since K has co-ordinates (a2, b2, c2), the line AK has equation c2y = b2z. The circumcircle Σ has equation a2yz + b2zx + c2xy = 0. (2.1) It follows that D has co-ordinates D(– a2, 2b2, 2c2). Similarly E has co-ordinates E(2a2, – b2, 2c2) and F has co-ordinates F(2a2, 2b2, – c2). The tangents at A, B and C have equations b2z + c2y = 0, c2x + a2z = 0 and a2y + b2x = 0 respectively. Those at B and C meet at the point U, which therefore has co-ordinates U(– a2, b2 , c2). Similarly V and W have co-ordinates V(a2, – b2, c2) and W(a2, b2, – c2). 3. The points X, Y, Z and the conic UVWXYZ The tangent to Σ at D has equation 4x/a2 + y/b2 + z/c2 = 0.
(3.1)
And the tangents to Σ at E and F have equations x/a2 + 4y/b2 + z/c2 = 0, x/a2 + y/b2 + 4z/c2 = 0.
(3.2) (3.3)
The tangents at E and F meet at X, which therefore has co-ordinates X(5a2, – b2, – c2). Similarly Y and Z have co-ordinates Y(– a2, 5b2, – c2) and Z(– a2, – b2, 5c2). We can now determine the equation of the conic that passes through the six points X, Y, Z, U, V, W. This has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (3.4) where u = 2b4c4, v = 2c4a4, w = 2a4b4, f = 3a4b2c2, g = 3a2b4c2, h = 3a2b2c4. In terms of u, v, w, f, g, h the centre of the conic with equation (3.4) has co-ordinates (x, y, z), where x = vw – gv – hw – f2 + fg + hf, y = wu – hw – fu – g2 +gh + fg, (3.5) 2 z = uv – fu – gv – h + hf + gh. The centre T of conic XYZUVW therefore has co-ordinates (x, y, z), where x = a2(3b2 + 3c2 – 5a2), 3
y = b2(3c2 + 3a2 – 5b2), z = c2(3a2 + 3b2 – 5c2).
(3.6)
4. The Brocard axis The Brocard axis is the line KO, where K is the symmedian point (a2, b2, c2) and O is the circumcentre of triangle ABC (a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)), and it therefore has equation (b2 – c2)x/a2 + (c2 – a2)y/b2 + (a2 – b2z/c2 = 0. (4.1) It may now be checked that T lies on the Brocard axis. 5. The sides of the triangles UVW and XYZ and the line LMN VW has equation c2y + b2z = 0,
(5.1)
and WU and UV have equations that may be written down from equation (5.1) by cyclic change of x, y, z and a, b, c. YZ has equation (3.1) and ZX and XY have equations (3.2) and (3.3). Sides VW and YZ meet at the point L, which therefore has co-ordinates L(0, – b2, c2). Similarly M and N have co-ordinates M(a2, 0, – c2) and N(– a2, b2, 0). It follows that L, M, N are collinear and line LMN has equation x/a2 + y/b2 + z/c2 = 0. (5.2) It is fairly easy to show that LMN is the polar of K with respect to Σ, and also the polar of K with respect to the conic UVWXYZ, and is perpendicular to the Brocard axis. 6. The sides of DEF The equation of EF is x/a2 = 2y/b2 + 2z/c2
(6.1)
The equations of FD and DE may now be written down from equation (6.1) by cyclic change of x, y, z and a, b, c. Note that the sides BC and EF pass through L. Similarly CA and FD pass through M and AB and DE pass through N. 4
7. The points L2, M2, N1, L1, N2, M1 and the conic passing through these points The point L2 = XY^WU and has co-ordinates (a2, 3b2, – c2). The point M2 = XY^VW and has co-ordinates (3a2, b2, – c2). The point N1 = ZX^VW and has co-ordinates (3a2, – b2, c2). The point L1 = ZX^UV and has co-ordinates (a2, – b2, 3c2). The point N2 = YZ^WU and has co-ordinates (– a2, 3b2, c2). The point M1 = YZ^UV and has co-ordinates (– a2, b2, 3c2). The conic through these six points has an equation of the form (3.4) with u = 2b4c4, v = 2c4a4, w = 2a4b4, f = 11a4b2c2, g = 11a2b4c2, h = 11a2b2c4. Using equations (3.6) the centre S of this conic has co-ordinates (x, y, z), where x = a2(11(b2 + c2) – 13a2), y = b2(11(c2 + a2) – 13b2), z = c2(11(a2 + b2) – 13c2).
(7.1)
It may now be checked that S lies on the Brocard axis and that LMN is the polar of K with respect to this conic also. 8. The points A', B', C', D', E', F' and the conic passing through these points The equations of the sides of triangle DEF are given in Section 6. The equations of AD, BE, CF are c2y = b2z, a2z = c2x, b2x = a2y respectively. The point A' = EF^AD and has co-ordinates (4a2, b2, c2). The point B' = FD^BE and has co-ordinates (a2, 4b2, c2). The point C' = DE^CF and has co-ordinates (a2, b2, 4c2). The point D' = BC^AD and has co-ordinates (0, b2, c2). The point E' = CA^BE and has co-ordinates (a2, 0, c2). The point F' = AB^CF and has co-ordinates (a2, b2, 0). It may now be shown that these six points lie on a conic, when expressed in the form (3.4) has u = b4c4, v = c4a4, w = a4b4, f = – a4b2c2, g = – a2b4c2, h = – a2b2c4. (8.1) This conic is the Brocard inellipse and it has centre R with co-ordinates (a2(b2 + c2), b2(c2 + a2), c2(a2 + b2)).
5
(8.2)
It may now be checked that R lies on the Brocard axis and that the polar of K with respect to this inellipse is the line LMN. 9. Review of points on the line LMN The following lines pass through L: BC, VW, YZ, EF, L1L2, B'C, E'F'. The first four of these have already been mentioned. The equations of the last three are: L1L2 2x/a2 = y/b2 + z/c2, (9.1) 2 2 2 B'C' 5x/a = y/b + z/c , (9.2) 2 2 2 E'F' x/a = y/b + z/c . (9.3) 2 2 It may be checked that L (0, – b , c ) lies on each of these lines. Seven similar lines pass through M and N. Points on LMN such as Q (a2, – 2b2, c2) have the following lines passing through it: AF, CD, BE (the same as VY), UZ, WX, L2N1, N2L1 showing it to be a vertex of a reverse perspective of triangles ABC, DEF and also UVW, XYZ. 10. The porisms Triangles ABC and DEF are inscribed in the circumcircle Σ and have sides that touch the Brocard inellipse at points A', B', C', D', E', F'. Thus a porism is established of triangles inscribed in Σ and circumscribing the Brocard inellipse. Triangles UVW and XYZ are inscribed in the conic centre T (see Section 3) and circumscribe (by construction) the circumcircle Σ, touching it at A, B, C, D, E, F. Thus a second porism is established.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article: CJB/117/2010
When the Cevians of triangle ABC meet the Circumcircle at D, E, F Christopher Bradley
N
U
F
31
A 21 E
G
V
23 Y
32 B
13
12
L
C
D
W
X
P
Q To R
M
Z
Fig. 1 When the Cevians through G meet the circumcircle and triangles are created
1
Abstract: In a triangle ABC when the Cevians through the centroid G meet the circumcircle Σ at points D, E, F and the tangents are drawn at the six points, two triangles are created. The intersections of the sides of the four triangles in the configuration have some interesting properties. 1. Introduction Fig. 1 is constructed as follows: Triangle ABC, its circumcircle and its centroid G are given. Lines AG, BG, CG meet the circumcircle again at points D, E, F respectively. Tangents to the circumcircle at A, B, C are drawn and they form a triangle XYZ, where X is the intersection of the tangents at B and C etc. Tangents at D, E, F are also drawn and they form the triangle UVW, where U is the intersection of tangents at E and F etc. We then prove: (i) that the points X, Y, Z, U, V, W lie on a conic, (ii) that the nine intersections of triangles ABC and DEF lie three on a line LMN and the other six 12 13 23 21 31 32 on a conic (ii) lines 12 21, 13 31, and 23 32 all pass through G, (iii) pairs of lines XY UV, YZ VW, ZX WU intersect on line LMN, which is therefore the Desargues’ axis of perspective of the two triangles and (iv) LMN is the polar of G with respect to circle ABCDEF, and with respect to the two conics just described. We use areal co-ordinates throughout with ABC as triangle of reference, though as David Monk has pointed out, property (ii) is a simple application of Pascal’s theorem. 2. Points D, E, F, X, Y, Z, U, V, W The equation of AG is y = z and this meets the circumcircle with equation a2yz + b2zx + c2xy = 0 at the point D, which therefore has co-ordinates D(– a2, b2 + c2, b2 + c2). Similarly E and F have co-ordinates E(c2 + a2, – b2, c2 + a2), F(a2 + b2, a2 + b2, – c2). The tangents at A, B, C have equations b2z + c2y = 0, c2x + a2z = 0,
a2y + b2x = 0.
(2.1)
The tangents at B and C meet at the point X, which therefore has co-ordinates X (– a2, b2, c2). Similarly Y and Z have co-ordinates Y(a2, – b2, c2), Z(a2, b2, – c2). The tangents at D, E, F have equations (b2 + c2)2x + a2b2y + c2a2z = 0, (c2 + a2)2y + b2c2z + a2b2x = 0, 2
(2.2) (2.3)
(a2 + b2)2z + c2a2x + b2c2y = 0. The tangents at E and F meet at U, which therefore has co-ordinates (x, y, z) where x = a4 + a2 (b2 + c2) + 2b2c2, y = – b2 (a2 + b2 – c2), z = – c2 (c2 + a2 – b2).
(2.4) (2.5) (2.6) (2.7)
V and W have co-ordinates which may be written down from Equations (2.5) – (2.7) by cyclic change of x, y, z and a, b, c. 3. The conic XYZUVW The conic through these six points has equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where u = 2a2b2c2(b2 + c2), v = 2a2b2c2(c2 + a2), w = 2a2b2c2(a2 + b2), f = a2{a4b2 + b4c2 + c4a2 + a2b4 + b2c4 + c2a4), g = b{a4b2 + b4c2 + c4a2 + a2b4 + b2c4 + c2a4), h = c{a4b2 + b4c2 + c4a2 + a2b4 + b2c4 + c2a4).
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
4. The lines EF, FD, DE and the line LMN Following on from Section 2 the line EF has equation – a2x + (c2 + a2)y + (a2 + b2)z = 0
(4.1)
The equations of the lines FD and DE follow from Equation (4.1) by cyclic change of x, y, z and a, b, c. The point L is the intersection of lines BC and EF and therefore has co-ordinates L(0, – 1/(c2 + a2), 1/(a2 + b2). Similarly M and N have co-ordinates M(1/(b2 + c2), 0, – 1/(a2 + b2)) and N(– 1/(b2 + c2), 1/(c2 + a2), 0). The points L, M, N are obviously collinear on the line with equation (b2 + c2)x + (c2 + a2)y + (a2 + b2)z = 0.
(4.2)
It may be checked that LMN is the polar of G with respect to the circumcircle and by construction is the Desargues’ axis of perspective of triangle ABC and DEF. It may also be checked that LMN is the polar of G with respect to the conic XYZUVW. Although we do not obtain the co-ordinates of P = YZ^VW, Q = ZX^WU, R = XY^UV Cabri indicates with 3
certainty that these points also lie on LMN, which is therefore the Desargues axis of perspective of triangles XYZ and UVW. 5. The points 12 23 31 21 32 13 and the conic through these points BC^FD = 12 which has co-ordinates (0, a2 + b2, b2). CA^DE = 23 which has co-ordinates (c2, 0, b2 + c2). AB^EF = 31 which has co-ordinates (c2 + a2, a2, 0). CA^EF = 21 which has co-ordinates (a2 + b2, 0, a2). AB^FD = 32 which has co-ordinates (b2, b2 + c2, 0). BC^DE = 13 which has co-ordinates (0, c2, c2 + a2). These six points lie on a conic of the form of Equation (3.1), where u = 2a2(b2 + c2), v = 2b2(c2 + a2), w = 2c2(a2 + b2), f = – {a4 + a2(b2 + c2) + 2b2c2}, g = – {b4 + b2(c2 + a2) + 2c2a2}, h = – {c4 + c2(a2 + b2) + 2a2b2}.
(5.1) (5.2) (5.3) (5.4) (5.5) (5.6)
It may now be checked that the line LMN is also the polar of G with respect to this conic. 6. The perspective of triangles 12 23 31 and 21 32 13 The equation of 23 32 is (b2 + c2) – b2y – c2z = 0.
(6.1)
It is immediate that 23 32 passes through G(1, 1, 1). Similarly lines 12 21 and 31 13 pass through G. As mentioned in Section 1 David Monk points out that this result also follows from Pascal’s theorem. Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
4
5
Article: CJB/2010/118
Two Triangles in Perspective and inscribed in a Conic Christopher Bradley
Q
Y to 18
6
3 To M
8
1
4
5
9
B'
V M' 49
38
D'
7 29
2
A
26
24
E C'
L
U L'
F
P
34 B E'
X R
57 37
27
D
C
67 A'
48
F'
N W Z
35
to 68
16 N'
Fig. 1 6 Pascal lines, 2 Steiner points, a Conic and many Points and Lines Abstract: Given two triangles ABC, DEF in perspective at P and inscribed in a conic Σ, a configuration is produced consisting of the 6 Pascal lines and 2 Steiner points arising. One of the Pascal lines has further properties, as it is not only a Pascal line, but also a Desargues’ axis of perspective and also the polar of P with respect to Σ. Further properties involving the nine lines AD, AE, AF, BD, BE, BF, CD, CE, CF and the tangents at the six vertices are deduced leading
1
to a conic, further properties of the key Pascal line and numerous sets of collinear points and concurrent lines. 1. Introduction Given a triangle ABC and a point P not on its sides lines AP, BP, CP are drawn to meet the circumconic Σ of ABC at points D, E, F respectively. The nine lines AD, AE, AF, BD, BE, BF, CD, CE, CF are drawn and in Fig. 1 these lines are marked 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. The six Pascal lines resulting from these nine lines are now drawn. These are Pascal line A: AE^BD, AF^CD, BF^CE, which is also the Desargues’ axis of perspective of triangles ABC and DEF; Pascal line B: AE^CD, AF^BD, BE^CF; Pascal line C: AD^BE, AF^CE, BF^CD; Pascal line D: AF^BE, AD^CE, BD^CF; Pascal line E: AE^BF, AD^CF, BD^CE; Pascal line F: AD^BF, AE^CF, BE^CD. We show that three of these pass through P and three through a point Q on Pascal line A. P and Q are therefore the 2 Steiner points of the configuration, which are common points of intersection of three Pascal lines. (In a complete configuration there are 60 Pascal lines and 20 Steiner points.) The point of this article is to study the interrelationship of the Pascal lines with the tangents to Σ at the vertices of the triangles ABC and DEF or if you like to study the interrelationship of the two properties: (i) ABC and DEF are inscribed in Σ and (ii) ABC and DEF are in perspective. First however we introduce our labelling of points in the configuration, which at first sight may appear confusing, but actually is critical for describing the configuration. We label AE^BD as 24 as AE is line 2 and BD is line 4 (see above); similarly AF^CD is point 37 and so on. Points in Fig.1 are labelled by this method. Now we describe some more key points, where tA for example means the tangent at A: (i) L is BC^tA and (as we shall prove) also lies on 35 29 and 34 P 27, (ii) M is CA^tB and also lies on 24 57 and 26 P 48, (iii) N is AB^tC and also lies on 48 27 and 67 P 38. We know from elementary theory that L, M, N are collinear. Now (iv) L' is EF^tD and also lies on 67 48 and 34 P 27; (v) M' is FD^tE and also lies on 38 27 and 26 P 48; (vi) N' is DE^tF and also lies on 35 16 and 67 P 38. Again L', M', N' are collinear. We prove that LMN and L'M'N' intersect at a point R on the key Pascal line A. Now tB and tC meet at A' which also lies on 48 57, tC and tA meet at B' which also lies on 38 29, tA and tB meet at C' which also lies on 35 26, tE and tF meet at D' which also lies on 38 29 B', t F and tD meet at E' which also lies on 34 16 C' and tD and tE meet at F' which also lies on 48 57 A'. The following collinearities also occur A'E' passes through 67 and 49, B'F' passes through 27 and 18 and C'D' passes through 26 and 35.
2
In the Sections that follow we give a co-ordinate treatment of the configuration using homogeneous projective co-ordinates with ABC as triangle of reference and P as unit point. 2. Points D, E, F and the nine lines AD, AE, AF, BD, BE, BF, CD, CE, CF We take P to be the unit point (1, 1, 1) and the circumconic Σ to have equation pyz + qzx + rxy = 0.
(2.1)
D is the point of intersection of the line y = z with the conic Σ and therefore has co-ordinates (– p, q + r, q + r). Similarly E and F have co-ordinates E(r + p, – q, r + p) and F(p + q, p + q, – r). Simple computations now provide the equations of the nine lines, which are: AD: y = z, AE: (r + p)y + qz = 0, AF: ry + (p + q)z = 0, BD: (q + r)x + pz = 0, BE: z = x, BF: rx + (p + q)z = 0, CD: py + (q + r)x = 0, CE: (r + p)y + qx = 0, CF: x = y.
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8) (2.9) (2.10)
3. The 18 points and the 6 Pascal lines We give the co-ordinates of the three points on a Pascal line and follow it up with the equation of the Pascal line itself. This is done six times, once for each Pascal line. 24: 37: 68: Pascal line A:
(p(r + p), q(q + r), – (r + p)(q + r)), (p(p + q), – (p + q)(q + r), r(q + r)), (– (p + q)(r + p), q(p + q), r(r + p). (q + r)x + (r + p)y + (p + q)z = 0.
(3.1)
27: 34: 59: Pascal line B:
(pq, – q(q + r), (r + p)(q + r)), (rp, (p + q)(q + r), – r(q + r)), (1, 1, 1). (q + r)x = ry + qz.
(3.2)
15: 38: 67:
(1, 1, 1), ((p + q)(r + p), – q(p + q), qr), (– p(p + q), (p + q)(q + r), rp). 3
Pascal line C:
(p + q)z = qx + py.
(3.3)
35: 18: 49: Pascal line D:
(r, – p – q, r), (– r – p, q, q), (p, p, – q – r). qx + ry + pz = 0.
(3.4)
26: 19: 48: Pascal line E:
((p + q)(r + p), qr, – r(r + p), (1, 1, 1), (–p(r + p), pq, (r + p)(q + r)). (r + p)y = rx + pz.
(3.5)
16: 29: 57: Pascal line F:
(– p – q, r, r), (q, q, – r – p), (p, p, – q – r). rx + py + qz = 0
(3.6)
The three Pascal lines B, C and E all pass through P(1, 1, 1). The three Pascal lines A, D, F all pass through a point Q with co-ordinates (p2 – qr, q2 – rp, r2 – pq). P and Q are the two Steiner points. It may now be verified that Pascal line A is the polar of P with respect to the conic Σ. 4. The lines EF, FD, DE and the points X, Y, Z From the co-ordinates of D, E, F in Section 2 we may obtain the equations of the lines EF, FD, DE. That of EF is – px + (r + p)y + (p + q)z = 0. (4.1) Those of FD and DE may be obtained from equation (4.1) by cyclic permutation of x, y, z and p, q, r. The point X = BC^EF then has co-ordinates X(0, – p – q, r + p). Similarly Y = CA^FD has coordinates Y(p + q, 0, – q – r) and Z = AB^ED has co-ordinates Z(– r – p, q + r, 0). It may now be checked that X, Y, Z all lie on Pascal line A, verifying that the Desargues’ axis of perspective coincides with Pascal line A. 5. The tangents at A, B, C, D, E, F and the points U, V, W The tangents at A, B, C have equations: tA: qz + ry = 0, tB : rx + pz = 0,
(5.1) (5.2) 4
tC :
py + qx = 0.
(5.3)
The tangents at D, E, F have equations: tD: (q + r)2x + pqy + rpz = 0, t E: pqx + (r + p)2y + qrz = 0, tF: rpx + qry + (p + q)2z = 0.
(5.4) (5.5) (5.6)
The tangent at A meets the tangent at D at the point U with co-ordinates U(p(q – r), – q(q + r), r(q + r)). Similarly points V and W have co-ordinates V(p(r + p), q(r – p), – r(r + p)) and W(– p(p + q), q(p + q), r(p – q)). It may now be checked that U, V, W also lie on Pascal line A. 6. The lines LMN and L'M'N' and the point R The line BC and the tangent at A meet at the point L with co-ordinates (0, q, – r). It may now be checked that L lies on the lines 35 29 and 34 P 27 with equations px + ry + qz = 0, (6.1) and ry + qz = (q + r)x. (6.2) Similarly M and N have co-ordinates M(– p, 0, r) and N(p, – q, 0). The line LMN clearly has the equation x/p + y/q + z/r = 0. (6.3) The line EF and the tangent at D have equations (r + p)y + (p + q)z = px, and (q + r)2x + pqy + rpz = 0.
(6.4) (6.5)
These lines meet at the point L' with co-ordinates L'(p(q – r), – rp – q(q + r), pq + r(q + r)). It may now be checked that L' lies on the lines D 67 48 and 34 P 27 which have equations rp(p + q)z + pq(r + p)y + (pq2 + pr2 + rq2 + qr2 + pqr)x = 0, (6.6) and ry + qz – (q + r)x = 0. (6.7) The co-ordinates of M' and N' may be written down from the co-ordinates of L' by cyclic change of x, y, z and p, q, r. The equation of the line L'M'N' may now be deduced and is p(q2 + qr + r2)x + q(r2 + rp + p2)y + r(p2 + pq + q2)z = 0. (6.8) The lines LMN and L'M'N' meet at the point R with co-ordinates R(p2(q – r), q2(r – p), r2(p – q)). It may now be checked that R also lies on Pascal line A. 5
7. The points A', B', C', D', E', F ' and the conic through these six points The tangents at B and C with equations given in Section 5 and the line 57 48 with equation (pq + r(q + r))x + p(p + r)y + prz = 0 (7.1) meet at the point A' with co-ordinates (– p, q, r). Similarly the tangents at A and C and the line 29 38 meet at the point B' with co-ordinates (p, – q, r). And similarly the tangents at A and B and the line 35 26 meet at the point C' with co-ordinates (p, q, – r). The tangents at E and F given in Section 5 and the line 38 29 with equation pqx + (p2 + pr + qr)y + q(p + q)z = 0
(7.2)
meet at the point D' with co-ordinates (x, y, z), where x = – (p2 + p(q + r) + 2qr), y = q(p + q – r), z = r(p – q + r).
(7.3) (7.4) (7.5)
The co-ordinates of E' and F’ may be written down from equations (7.3) – (7.5) by cyclic change of x, y, z and p, q, r. The point E' is the intersections of the tangents at F and D and the line 34 16. The point F' is the intersection of the tangents at D and E and the lie 48 57. It can now be shown that the points A', B', C', D', E,' F' lie on a conic with equation of the form ux2+ vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (7.6) with u = 2pqr(q + r), (7.7) v = 2pqr(r + p), (7.8) w = 2pqr(p + q), (7.9) 2 2 2 2 2 2 f = p(p q + p r + q r + q p + r p + r q), (7.10) 2 2 2 2 2 2 g = q(p q + p r + q r + q p + r p + r q), (7.11) 2 2 2 2 2 2 h = r(p q + p r + q r + q p + r p + r q). (7.12) 8. Three unexplained collinearities The equation of C'D' is rpx + r(r + p)y + (p2 + pq + pr)z = 0.
(8.1)
It may now be shown that points 26 and 35 lie on this line. Similarly B'F' passes through points 18 and 27, and A'E' passes through the points 49 and 67.
6
Flat 4 Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
7
Article: CJB/2010/119
The Nine-point Conic and a Pair of Parallel Lines Christopher Bradley Abstract: An affine transformation of the nine-point circle leads to the nine-point conic. The orthocentre transforms into a general point P which is collinear with the centre of the nine-point conic and the centroid. We prove that the polar of P with respect to the nine-point conic is parallel to the Desargues’ axis of perspective of the triangle and the image of the orthic triangle.
A
U
N
M
G T B
V'
Q S
V L
P R
W A'
C B'
U'
R' To W'
S'
C' W' T'
Fig.1 The nine-point conic 1. Introduction
1
A triangle ABC is given and also a general point P not on the sides of ABC. The Cevians AP, BP, CP meet the sides of ABC at points R, S, T. The midpoints of the sides of ABC are denoted by L, M, N. The conic RSTLMN is the nine-point conic Σ and it passes through the midpoints U, V, W of AP, BP, CP respectively. The points R', S', T' on BC, CA, AB are the harmonic conjugates of R, S, T respectively. It is verified that this is the Desargues’ axis of perspective of triangles ABC and PQR (as is always the case). The polar of P with respect to the conic Σ meets the sides BC, CA, AB respectively at A', B', C'. It is proved that A'B'C' is parallel to R'S'T'. The point U' = AR^BS' is defined, with V' and W' similarly defined. The Desargues’ axis of perspective of triangle ABC and U'V'W' is also the line R'S'T. We prove results, as necessary, using areal co-ordinates with ABC as triangle of reference. 2. The conic LMNRST The points L, M, N have co-ordinates L(0, 1, 1), M(1, 0, 1), N(1, 1, 0). Suppose P has coordinates (l, m, n). Then R, S, T have co-ordinates R(0, m, n), S(l, 0, n), T(l, m, 0). It may now be shown that the conic Σ passing through these six points has equation mnx2 + nly2 + lmz2 – l(m + n)yz – m(n + l)zx + n(l + m)xy = 0. (2.1) Σ meets AP again at the point U with co-ordinates U(2l + m + n, m, n). Similarly it passes through V(l, l + 2m +n, n) and W(l, m, l + m + 2n). It is easy to see that U, V, W are the midpoints of AP, BP, CP respectively. This does, of course follow from the fact that the ninepoint conic arises from the nine-point circle by means of an affine transformation, which preserves ratios of line segments. 3. The centre Q of Σ and the collinearity of P, Q, G where G is the centroid of ABC The centre Q of the conic Σ has co-ordinates Q(2l + m + n, l + 2m + n, l + m + 2n). It may be checked that G, Q, P are collinear and that GQ = (1/3)QP, which reflects the fact that on the Euler line GN = (1/3)NH, where N is the centre of the nine-point circle and H is the orthocentre. 4. The polar of P with respect to Σ The equation of the polar of P with respect to Σ is mn(m + n)x + nl(n + l)y + lm(l + m)z = 0.
(4.1)
This meets BC at A' with co-ordinates A'(0, – m(l + m), n(n + l)). Points B', C' on CA, AB respectively have co-ordinates B'(l(l + m), 0, – n(m + n)), C'(– l(n + l), m(m + n), 0). 5. The points R', S', T' and the result that A'B'C' is parallel to R'S'T'
2
The equation of the line ST is nly + lmz = mnx.
(5.1)
This meets the side BC at the point R' with co-ordinates R'(0, – m, n). Points S', T' on CA, AB respectively have co-ordinates S'(l, 0, – n), T'(– l, m, 0). The points R', S', T' are harmonic conjugate of R, S, T respectively with respect to B, C and C, A and A, B. R'S'T' is, of course a straight line, the transversal associated with the Cevian point P, and again is well-known to be the Desargues’ axis of perspective of triangles ABC and RST. The equation of R'S'T' is found to be mnx + nly + lmz = 0.
(5.2)
The point of intersection of A'B'C' and R'S'T' has co-ordinates (l(m – n), m(n – l), n(l – m)) which lies on the line at infinity x + y + z = 0 and hence A'B'C' and R'S'T' are parallel. 6. The points U', V', W' The lines AR and BS' meet at the point U' with co-ordinates U'(– l, m, n). Similarly BS and CT' meet at V'(l, – m, n) and CS and AR' meet at W'(l, m, – n). The equation of the line V'W' is therefore ny + mz = 0, which meets BC at R'. It follows that R'S'T' is also the Desargues’ axis of perspective of triangles ABC and U'V'W'.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/120
An Important Line through the Centre of a Cevian Inellipse Christopher Bradley Abstract: A Cevian inellipse is one that touches the sides of a triangle internally at points that are the feet of Cevian lines. It is shown in this short article that the line through the Cevian point and the centre of a Cevian inellipse contains other significant points.
A
N
M
O
Q
B
K F
L
C
Fig. 1 1. Introduction Let ABC be a triangle and P an internal point. Cevian lines through P meet the sides BC, CA, AB at L, M, N respectively. The conic touching BC at L, CA at M and AB at N is what we term a Cevian inellipse. If Q is the centre of a Cevian inellipse we show that other significant points inevitably lie on the line PQ. Examples when the Cevian point is the symmedian point K and the incentre I are given. 2. The Cevian Inellipses We consider first the case when there are two Cevian points P and T with co-ordinates P(l, m, n) and T(u, v, w).
1
The feet of these Cevians are L (0, m, n), M(l, 0, n), N(l, m, 0), U(0, q, r), V(p, 0, r), W(p, q, 0).
A
N M Mi Q
B
L
I K
C
Fig. 2 It is straightforward to show that the equation of the conic through the six points L, M, N, U, V, W is x2/lp + y2/mq + z2/nr – (mr + nq)yz/(mnqr) – (lr + np)zx/(lnpr) – (lq + mp)xy/(lmpq) = 0. (2.1) Putting p = l, q = m, r = n we obtain the equation of the Cevian inellipse touching BC at L, CA at M, AB at N. Its equation, therefore, is x2/l2 + y2/m2 + z2/n2 – 2yz/mn – 2zx/nl – 2xy/lm = 0. (2.2) In Fig.1 we show this conic when P is the symmedian point K and in Fig.2 we show this conic when P is the incentre I. When P is at K we have the Brocard inellipse. 3. The centre Q and the line PQ Using standard formulae the co-ordinates of the centre Q of this conic are (x, y, z), where x = l(m + n), y = m(n + l), z = n(l + m) (3.1) The co-ordinates of P are P(l, m, n) and so the line PQ has the equation mn(m – n)x + nl(n – l)y + lm(l – m)z = 0. 2
(3.2)
It may be verified that PQ not only passes through P(l, m, n) and Q(l(m + n), m(n + l), n(l + m) but also the points with co-ordinates (l2, m2, n2) and (l(m + n – l), m(n + l – m), n(l + m – n)) and indeed many other simple combinations of l, m, n. 4. Two examples In Fig. 1 we show the case when P is at the symmedian point K, The centre Q has coordinates Q(a2(b2 + c2), b2(c2 + a2), c2(a2 + b2)), the two additional points being the fourth power point F(a4, b4, c4), and the circumcentre O(a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2). The line is, of course the Brocard axis and the inellipse is the Brocard inellipse. In Fig.2 when P is at the incentre I then Q has co-ordiantes Q(a(b + c), b(c + a), c(a + b)) and the additional points have co-ordinates (a2, b2, c2), the symmediam point K and the Mittenpunkt Mi with co-ordinates (a(b + c – a), b( c + a – b), c(a + b – c)). In the Encyclopedia of Triangle Centres (Kimberling) a vast number of collinearities are given, including these. It is hoped, however, that the generalization in this article provides a geometrical reason for many of these collinearities. Lists, brilliant and vast as they may be, provide no geometrical insight whatsoever. In addition it may be noted that once you have a circumcircle and a Cevian inellipse you automatically create a porism of triangles with vertices on the circumcircle that circumscribe the inellipse. The Brocard porism, when P is at K and Poncelet’s porism when P is at Gergonnes’ point are the best known.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2010/121
Some Lines through the Centroid G Christopher Bradley Abstract: Using areal co-ordinates we catalogue some lines through the centroid G of a triangle ABC. Co-ordinates of points and equations of lines are given. 1. The Symmedian line
A C'
B'
N
Ka
M
KmG
K
L
B
C
G(1,1,1) K (a^2,b^2,c^2)
Km (b^2+c^2,..)
A'
Ka (b^2+c^2 -a^2,..)
Fig. 1 The equation of the line through the three symmedian points is (b2 – c2)x + (c2 – a2)y + (a2 – b2)z = 0. (1) Km is the symmedian of the Medial triangle and Ka is the symmedian of the Anticomplementary triangle. Ka has co-ordinates (b2 + c2 – a2, c2 + a2 – b2, a2 + b2 – c2) and Km has co-ordinates (b2 + c2, c2 + a2, a2 + b2). K, of course, has co-ordinates (a2, b2, c2).
1
2. The Euler line
A
deL
O
Sch
G N H
B
C
Fig. 2 The Centroid G(1, 1, 1), the Circumcentre O(a2(b2 + c2 – a2), .., ..), the Orthocentre H(1/(b2 + c2 – a2), .., ..), the Nine-Point Centre N(a2(b2 + c2) – (b2 – c2)2, .., ..), deLongchamps point deL(– 3a4 + 2a2(b2 + c2) + (b2 – c2)2, .., ..), Schiffler’s point Sch(a/(cos B + cos C), .., ..). The equation of the Euler line is (b2 + c2 – a2)(b2 – c2)x + (c2 + a2 – b2)(c2 – a2)y + (a2 + b2 – c2)(a2 – b2)z = 0.
(2)
Schiffler’s point is the intersection of the Euler’s line of ABC, IBC, ICA, IAB, where I is the incentre.
2
3. The Incentre Centroid line
A
Na
Sp G I
C
B
Fig. 3 This line passes through G(1, 1, 1), I(a, b, c), the Spieker Centre Sp(b + c, c + a, a + b) and Nagel’s point (b + c – a, c + a – b, a + b – c). It is the analogue of the Euler line when O is replaced by I. It is also the analogue of the Symmedian line when the symmedian point is replaced with the incentre. The Spieker Centre is the centroid of the three sides of the triangle with no internal material present. It is also the incentre of the medial triangle, as its co-ordinates imply. These last two properties are, in fact, the same property. The equation of this line is (b – c)x + (c – a)y + (a – b)z = 0 (3)
3
4. The Mittenpunkt line
A
F M N Mi
G Ni
Mi*
E
Ge
B
L
D
C
Fig. 4 This line contains the Mittenpunkt Mi(a(b + c – a), b(c + a – b), c(a + b – c)), the isogonal conjugate of the Mittenpunkt Mi*(a/(b + c – a), b/(c + a – b), c/(a + b – c)), Gergonne’s point Ge(1/(b + c – a), 1/(c + a – b), 1/(a + b – c)), the centroid G(1, 1, 1), the centre of the Nine-Point Conic (with Cevian points G and Ge) Ni(a(b + c) – (b – c)2, b(c + a) – (c – a)2, c(a + b) – (a – b)2) As it is an analogue of the Euler line (when Mi replaces O) one has GGe = 2MiG and MiNi = NiGe. It also has many other less important points lying on it. Its equation is (b + c – a)(b – c)x + (c + a – b)(c – a)y + (a + b – c)(a – b)z = 0.
(4)
It should be noted that we can obtain countless lines through G by using a point P other than G (not on the sides of ABC) and drawing the Cevians APD, BPE, CPF, constructing the conic DEFLMN and finding its centre Q. Then PGQ is a straight line and QP = 3GQ.
4
5. The Centroid Feuerbach line
A
Feu
O
IS
I
ES
G
C
B
Fig. 5a In Fig. 5 IS and ES are the centres of similitude of the circumcircle and the incircle and so lie on the line IO and are harmonic conjugates with respect to O and I. IS has co-ordinates (a2(b + c – a), b2(c + a – b), c2(a + b – c)) and ES has co-ordinates (a2/(b + c – a), b2/(c + a – b), c2(a + b – c)). The line OI IS ES has equation bc(b – c)(b + c – a)x + ca(c – a)(c + a – b)y + ab(a – b)(a + b – c)z = 0. (5a) IS also lies on the line joinin G to Feurbach’s point Fe, the point where the nine-point circle touches the incircle. Feurbach’s co-ordinates are ((b + c – a)(b – c)2, (c + a – b)(c – a)2, (a + b – c)(a – b)2). The line G Fe has equation (b – c)(a(b + c) – b2 – c2)x + (c – a)(b(c + a) – c2 – a2)y + (a – b)(c(a + b) – a2 – b2)z = 0. (5b)
5
A
Feu
O IS
I ES G FeH N
C
B
Fig. 5b It is also the case that G ES passes through the harmonic conjugate FeH of Feu with respect to I (the incentre) and N(the nine-point centre). The co-ordinates of FeH are (bc(b + c)2/(b + c – a), ca(c + a)2/ (c + a – b), ab(a + b)2/(a + b – c)) and the line G FeH ES has equation (b + c – a)(b – c)(a(b + c) + b2+ c2)x + (c + a – b)(c – a)(b(c + a) + c2 + a2)y + (a + b – c)(a – b)(c(a + b) + a2 + b2)z = 0. (5c)
Flat 4 Terrill Court 12-14, Apsley Road BRISTOL BS8 2SP
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Article: CJB/2010/122
Some lines through the Incentre I Christopher Bradley
Abstract: Using areal co-ordinates we catalogue some lines through the incentre I of a triangle ABC. Co-ordinates of points and equations of lines are given. I3
A
I2 N E
F
M
Q Mi
D L
B
K I
C
I1
Fig. 1 We illustrate above the line through I(a, b, c) passing through the symmedian point K(a2, b2, c2). It also passes through the Mittenpunkt, with co-ordinates (a(b + c – a), b(c + a – b), c(a + b – c)), which is the symmedian point of the triangle of excentres, and Q, the Cross point of Incentre and 1
Centroid (X37 in the Encyclopaedia of Triangle Centres), which has co-ordinates (a(b + c), b(c + a), c(a + b)). If you take the triangles ANM, BNL, CLM and their centroids, then the lines through A, B, C through these respective centroids are concurrent at Q. However it is better thought of as the centre of the inellipse touching the sides of ABC at L, M, N, where L, M, N are the feet of the Cevians through I. This is the prime example of an infinite number of collinearities through P(l, m, n), R(l 2. m2, n2), S(l(m + n – l), m(n + l – m), n(l + m – n)) and Q(l(m + n), m(n + l), n(l + m)), where P is any general point and Q is the centre of the inconic touching the sides of ABC at the feet of the Cevians through P. The equation of all such lines is mn(m – n)x + nl(n – l)y + lm(l – m)z = 0. (1) Another example is the Brocard axis when l = a2, m = b2, n = c2 when P = K, R is F, the fourth power point, Q is the centre of the Brocard inellipse and S is the circumcentre O.
A
deL
I Ge
C
B
Fig. 2 If l = 1/(b + c – a), m = 1/(c + a – b), n = 1/(a + b – c) then (l, m, n) is Gergonne’s point, (l(m + n), m(n + l), n(l + m)) is the incentre and IGe is mn(m – n)x + nl(l – m)y + lm(l – m)z = 0
2
or (b + c – a)2(b – c)x + (c + a – b)2(c – a)y + (a + b – c)2(a – b)z = 0. Clearly (l2, m2, n2) lies on this line, but currently I have no reference to its position. deL(– 3a4 + 2a2(b2 + c2) + (b2 – c2)2, .., ..) also lies on this line. Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article: CJB/2010/123 P1 Properties of the Intangential Triangle
Christopher Bradley F2
I3
To P1 E3 Q3
A'
Q2
To I3
A W
N' M
N
F3
I2
V
F Z I
P H
E2 Y E M'
To Q2
B' F1
B
L D1
X
D
P2
C
E1
L' D3 To I1
U
C' To Q1 D2 P3
I1
Q1
Fig. 1 The Intangential Configuration Abstract: Triangle ABC and its incircle and ex-circles are given. Common tangents to the incircle and an ex-circle are four in number; three being the sides of ABC and the fourth is called an intangent. There are three intangents, one for each ex-circle and these form a triangle A'B'C' called the intangential triangle. Properties of the intangential triangle are studied using areal coordinates with ABC as triangle of reference. By construction its incircle is the same as that of triangle ABC. The triangle of its excentres has sides parallel to those of ABC. 1
1. Introduction First we introduce the points in Fig. 1 above in the order they were constructed. An arbitrary triangle ABC is given. The incircle, centre I touches BC, CA, AB respectively at D, E, F respectively. The ex-circles are drawn, with centres I1, I2, I3 for the circles opposite A, B, C respectively. Points of contact of the ex-circle opposite A are labeled D1, D2, D3, subscripts 1, 2, 3 indicating tangency on BC, CA, AB respectively. Similarly points of contact of the ex-circle opposite B are labeled E1, E2, E3 and points of contact of the ex-circle opposite C are labeled F1, F2, F3. Points X, Y, Z are the intersections of AII1, BII2, CII3 respectively with BC, CA, AB. If one takes the incircle and the ex-circle opposite A, then they already have three common tangents drawn, namely the sides BC, CA, AB of triangle ABC. They have a fourth common tangent called an intangent, which is the line through X touching the incircle at L and the excircle at L'. The fact that the intangent passes through X has, of course, to be established. Points M, M' are the points where the second intangent touches the incircle and the ex-circle opposite B and points N, N' are the points where the third intangent touches the incircle and the ex-circle opposite C. The second and third intangents meet at A', and B' and C' are similarly defined. Triangle A'B'C' is called the intangential triangle Δ. By construction Δ has the same incircle as triangle ABC with contact points L, M, N. Its ex-circles are also drawn with centres at Q1, Q2, Q3. The lines D2D3, E3E1, F1F2 form the triangle P1P2P3. A number of properties are proved in the following sections: (i) the points D2, D3, E3, E1, F1, F2 lie on a conic; (ii) the points D, D1, E, E2, F, F3 lie on a conic; (iii) the points A, A', B, B', C, C' lie on a conic; (iv) line Q2Q3 passes through A', line Q3Q1 passes through B' and line Q1Q2 passes through C'; (v) triangle LMN has its sides parallel to those of ABC; (vi) triangle Q1Q2Q3 also has its sides parallel to those of ABC; (vii) A', I, D, Q1 are collinear on a line perpendicular to BC and similarly for lines B'IEQ2 and C'IFQ3; (viii) the incentre I lies on the conic AA'BB'CC'. A number of properties were tested and proved not to be true. For example L, L', M, M', N, N' do not lie on a conic and (see Fig. 1) UVW is not the incircle of triangle P1P2P3. 2. Well known facts about the incircle and the ex-circles The incircle has equation (b + c – a)2x2 + (c + a – b)2y2 + (a + b – c)2z2 – 2(c + a – b)(a + b – c)yz – 2(a + b – c)(b + c – a)zx – 2(b + c – a)(c + a – b)xy = 0.
(2.1)
It touches BC at the point D with co-ordinates D(0, a + b – c, c + a – b). Similarly the points E and F have co-ordinates E(a + b – c, 0, b + c – a) and F(c + a – b, b + c – a, 0).
2
The ex-circle opposite A has equation (a + b + c)2x2 + (a + b – c)2y2 + (c + a – b)2z2 – 2(c + a – b)(a + b – c)yz + 2((a + b + c)(c + a – b)zx + 2(a + b + c)(a + b – c)xy = 0. (2.2) It touches BC at the point D1 with co-ordinates D1 (0, c + a – b, a + b – c). Similarly E2 and F3 have co-ordinates E2(b + c – a, 0, a + b – c) and F3(b + c – a, c + a – b, 0). The ex-circle opposite A meets the side CA at D2 with co-ordinates D2(b – c – a, 0, a + b + c) and it meets the side AB at D3 with co-ordinates D3(c – a – b, a + b + c, 0). Similarly E3 has coordinates E3(a + b + c, c – a – b, 0) and E1 has co-ordinates E1(0, a – b – c, a + b + c). And again F1 has co-ordinates (0, a + b + c, a – b – c) and F2 has co-ordinates F2(a + b + c, 0, b – c – a). 3. The intangents and the intangential triangle The proposal is that there is a line through the point X, the intersection of II1 and BC that touches the incircle at a point L, and which touches the ex-circle opposite A at a point L'. What we shall do is to determine the line through X that touches the incircle and then show that the same line also touches the ex-circle opposite A. How we determined the line XL is immaterial (it took some time and care) but once obtained it is simply a matter of checking that it has the desired properties. We assert that the line XL has equation bc(c – b)(b + c – a)x – (c – a)2(b + c – a)(cy – bz) = 0.
(3.1)
and that the point L has co-ordinates L((b + c – a)(b – c)2, b2(c + a – b), c2(a + b – c)). It may now be shown that L' has co-ordinates L'(–(a + b + c)(b – c)2, b2(b + a – c), c2(c + a – b)). Points M, M', N, N' have co-ordinates which may be written down from those of L, L' by cyclic change of x, y, z and a, b, c. Similarly the equations of MYM' and NZN' may be written down from equation (3.1) by cyclic change of x, y, z and a, b, c. These three lines are called the intangents. LXL' and MYM' meet at the point C' with co-ordinates C'(a(b + c – a)(b – c), b(c + a – b)(a – c), c2(a + b – c)). Point A' and B' have co-ordinates A'(a2(b + c – a), b(c + a – b)(c – a), c(a + b – c)(b – a)), and B'(a(b + c – a)(c – b), b2(c + a – b), c(a + b – c)(a – b)). These points A', B', C' are the vertices of the intangential triangle.
3
(3.2)
(3.3) (3.4)
4. The three conics It may be shown that points A, A', B, B', C, C', I lie on a conic. Its equation is fyz + gzx + hxy = 0,
(4.1)
where f = a2(b – c)(a2 + b2 + c2 +2bc – 2ca – 2ab),
(4.2)
and g, h may be written down from Equation (4.2) by cyclic change of a, b, c. It may also be shown, as is well known, that points D, E, F, D1, E2, F3 lie on a conic with equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (4.3) where u = v = w = (b + c – a)2(c + a – b)2(a + b – c)2, f = – (b + c – a)2(c + a – b)(a + b – c)(a2 + (b – c)2),
(4.4) (4.5)
and where g, h may be written down from Equation (4.5) by cyclic change of a, b, c. It may also be shown, as again is well known that points D2, D3, E3, E1, F1, F2 lie on a conic with equation of the form (4.3), where u = v = w = ((b + c)2 + a2)((c + a)2 + b2)((a + b)2 + c2), (4.6) f = – (a2 + b2 + c2 + 2ab)(a2 + b2 – c2 + 2bc) (a2 – b2 + c2 + 2ca), (4.7) and where g, h may be written down from Equation (4.7) by cyclic change of a, b, c. 5. The points P1, P2, P3, U, V, W The lines D2D3, E3E1, F1F2 form a triangle with vertices P1, P2, P3 whose co-ordinates are P1(2a(b + c), c2 – a2 – b2, b2 – c2 – a2), P2(c2 – a2 – b2, 2b(c + a), a2 – b2 – c2), P3(b2 – c2 – a2, a2 – b2 – c2, 2c( a + b)). Curiously enough the centre of the circle P1P2P3 is the orthocentre of triangle ABC. The line AII1 meets D2D3 at the point U with co-ordinates U(2bc – ca – ab – b2 – c2, b(a + b + c), c(a + b + c)). Points VW have co-ordinates that may be written down from those of U by cyclic change of x, y, z and a, b, c. 6. The triangle LMN
4
The line MN has equation (a2 – ab – ac + 2bc)x – a(b + c – a)(y + z) = 0.
(6.1)
The lines NL and LM follow from Equation (6.1) by cyclic change of a, b, c and x, y, z. MN is parallel to BC, so the triangle LMN is similar to triangle ABC by a rotation about the incentre I through an angle of 180o followed by an enlargement by a factor that is triangular dependent. 7. The triangle Q1Q2Q3 We omit the working which is technically quite difficult but concludes with the co-ordinates of the ex-centres of triangle A'B'C' labeled Q1, Q2, Q3. The co-ordinates of Q1 are (x, y, z) where x = – a( b + c – a)(a2 + (b – c)2), (7.1) 2 2 2 y = b(c + a – b)(a + b – c ), (7.2) 2 2 2 z = c(a + b – c)(a – b + c ). (7.3) The co-ordinates of Q2, Q3 follow from Equations (7.1) – (7.3) by cyclic change of x, y, z and a, b, c. The equation of Q2Q3 is (ab + ac – b2 – c2)x + a(a + b – c)(y + z) = 0,
(7.4)
and the equations of Q3Q1 and Q1Q2 follow from Equation (7.4) by cyclic change of x, y, z and a, b, c. Q2Q3 is also parallel to BC, so the triangle Q1Q2Q3 is similar to triangle ABC by a rotation about a point P through an angle of 180o followed by an enlargement by a factor that is triangular dependent. Cabri indicates that P is X33 in Kimberling’s Encyclopedia of Triangle Centres. It lies on IH and is the perspector of the intangential and orthic triangles. A'Q1, B'Q2, C'Q3 are, of course, concurrent at the incentre I, since I is also the incentre of triangle A'B'C'.
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Article: CJB/2010/124
Two Cevian Points Collinear with a Vertex and Thirteen Conics Christopher Bradley Abstract: Given triangle ABC, a point P on BC and lines LMP and UVP with L, U on AB and M, V on CA, then two Cevian points may be constructed collinear with vertex A. Two further Cevian points occur and the resulting configuration of points and lines results in the appearance of thirteen conics and two harmonic ranges.
A
U
L
V Q
F S
M
T R
B
X
E
D
Y
C
Fig. 1 Two Cevian points Q, R collinear with A and the thirteen resulting conics 1. Introduction
1
P
The construction of the above figure is as follows. Triangle ABC is given together with a point P on BC (not B or C). Two points U and L on AB are chosen and lines UVP, LMP are drawn with V, M on CA. The following lines are now drawn: BM, BV, CU, CL. These define four points, BV^CU = Q, BM^CL = R, BV^CL = S and BM^CU = T. The line QR is shown to pass through A and the harmonic conjugate D of P relative to B and C. The line ST meets AB at F and CA at E and is shown to pass through P. Lines AS, AT meet BC at X, Y respectively and X, Y are also harmonic conjugates of D, P. The following thirteen conics are now shown to exist: XYLMUV, XYLMEF, XYEFUV, UVLMST, PQRLVX, PQRMUY, DLSPVA, DMTPUA, LMUVD touching BC at D, UVEFD touching BC at D, LMEFD touching BC at D, STXYUV and STXYLM. The four points Q, R, S, T are Cevian points and two of them Q, R are collinear with a vertex. This article highlights what to expect when a pair of Cevian points are collinear with a vertex, and at the same time, though we make no analysis of this context at present, when three related complete quadrilaterals exist with points B, C, P in common. In what follows areal co-ordinates are used with ABC as triangle of reference. There appears to be no simplification to be had by using homogeneous projective co-ordinates. 2. The lines and points We take P to have co-ordinates (0, p, 1 – p), M to have co-ordinates M(1 – m, 0, m) and V to have co-ordinates V(1 – v, 0, v). The equation of PLM is therefore mpx + (m – 1)(p – 1)y + (m – 1)pz = 0, (2.1) with L having co-ordinates L(– (p – 1)(m – 1), pm, 0). Similarly the equation of PUV is vpx + (v – 1)(p – 1)y + (v – 1)pz = 0, (2.2) with U having co-ordinates U(– (p – 1)(v – 1), pv, 0). The equations of BV, BM, CU, CL are as follows: BV: vx + (v – 1)z = 0, BM: mx + (m – 1)z = 0. CU: (p – 1)(v – 1)y + pvx = 0, CL: (p – 1)(m – 1)y + pmx = 0. The co-ordinates of the Cevian points Q, R, S, T may now be determined and are 2
(2.3) (2.4) (2.5) (2.6)
Q = BV^CU: R = BM^CL: S = BV^CL: T= BM^CU:
((1 – v)(p – 1), vp, v(p – 1)), ((1 – m)(p – 1), mp, m(p – 1)), ((1 – v)(m – 1)(p – 1), pm(v – 1), v(m – 1)(p – 1)), ((1 – v)(m – 1)(p – 1), pv(m – 1), m(v – 1)(p – 1)).
The equation of ST may now be obtained and is p(2mv – m – v)x + (m – 1)(v – 1)(p – 1)y + (m – 1)(v – 1)pz = 0. Result 1 (i) (ii) (iii)
Result 2 (i) (ii) (iii) (iv) (v) (vi)
(2.7) (2.8) (2.9) (2.10)
(2.11)
ST passes through P: when x = 0, y = p and z = (1 – p). ST meets CA at E with co-ordinates E((1 – v)(m – 1), 0, 2mv – m – v). ST meets AB at F with co-ordinates F((1 – v)(m – 1)(p – 1), p(2mv – m – v), 0).
The equation of QR is (p – 1)y = pz. (2.12) QR passes through A(1, 0, 0). AQR meets BC at D(0, p, p – 1), the harmonic conjugate of P with respect to B and C. AS meets BC at X with co-ordinates X(0, mp(v – 1), v(m – 1)(p – 1)). (2.13) AT meets BC at Y with co-ordinates Y(0, vp(m – 1), m(v – 1)(p – 1)). (2.14) D is the harmonic conjugate of P with respect to X and Y. The proof is left to the reader.
3. The Thirteen Conics The equation of any conic may be put in the form ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0. (3.1) Here a, b, c are not the side lengths, but constants to be determined. For each conic we give the values of a, b, c, f, g, h in terms of p, m, v. Conic XYLMUV a = 2m2p2v2(m + v – 2mv), b = 2mv(1 – v)(m – 1)(p – 1)2(2mv – m – v), c = 2mv(1 – v)(m – 1)p2(2mv – m – v), f = p(p – 1)(2mv – m – v)(2m2v2 – 2mv(m + v) + m2 + v2), g = – mvp2(2mv – m – v)2, h = – p(p – 1)mv(2mv – m – v)2. Conic XYLMEF a = 2m2p2v(2mv – m – v), b = 2mv(m – 1)2 p – 1)2(v – 1), c = 2mvp2(m – 1)2(v – 1), 3
f = p(1 – p)(m – 1)(2m2v2 – 2mv(m + v) + m2 + v2), g = mvp2(m – 1)(3mv – 2m – v), h = mv(m – 1)p(p – 1)(3mv – 2m – v). Conic XYUVEF Exchange m and v in the expressions for conic XYLMEF. Conic UVLMST a = 2mvp2, b = 2(m – 1)(v – 1)(p – 1)2, c = 2(m – 1)(v – 1)p2, f = p(m – 1)(v – 1)(p – 1), g = p2(2mv – m – v), h = p(p – 1)(2mv – m – v). Conic PQRVLX a= 4mvp2(m – v), b = 2v(v – 1)(m – 1)2(p – 1)2, c = 2m(1 – m)(v – 1)2p2, f = p(p – 1)(m – v)(m – 1)(v – 1), g = m(1 – v)p2(mv – 2m + v), h = p(p – 1)v(m – 1)(mv + m – 2v). Conic PQRUMY Exchange m and v in the expressions for conic PQRVLX. Conic DLSPVA a = 0, b = 2(m – 1)(v – 1)(p – 1)2, c = 2p2(1 – v)(m – 1), f = 0, g = p2v(1 – m), h = p(1 – p)m(1 – v). Conic DMTPUA Exchange m and v in the expressions for conic DLSPVA. Conic LMUVD touching BC at D a = 2mvp2, b = 2(m – 1)(v – 1)(p – 1)2, c = 2(m – 1)(v – 1)p2, f = 2p(1 – p)(v – 1)(m – 1), g = p2(2mv – m – v), h = p(p – 1)(2mv – m – v). Conic UVEFD touching BC at D a = 2p2v(2mv – m – v), b = 2(m – 1)(v – 1)2(p – 1)2, c = 2p2(m – 1)(v – 1)2, f = 2p(1 – p)(m – 1)(v – 1)2, g = p2(v – 1)(3mv – m – 2v), h = p(p – 1)(v – 1)(3mv – m – 2v). Conic LMEFD touching BC at D Exchange m and v in the expressions for conic UVEFD. Conic STXYUV a = 2p2v3(m – 1)(2mv – m – v), b = 2mv(1 – v)3(m – 1)(p – 1)2, c = 2mvp2(1 – v)3(m – 1), f = p(p – 1)(v – 1)2(2m2v2 – 2mv(m + v) + m2 + v2), g = p2v3(m – 1)2(v – 1), h = pv3(m – 1)2(p – 1)(v – 1). 4
Conic STXYLM Exchange m and v in the expressions for conic STXYUV.
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Article: CJB/2011/125
Properties of the Extangential Triangle Christopher Bradley Abstract: The external common tangents to the three ex-circles form the extangential triangle A'B'C'. Triangles ABC and A'B'C' are in perspective with perspector the orthocentre of the intouch triangle. The axis of perspective plays a prominent part in the configuration. Triangle A'B'C' and the orthic triangle are homothetic through the Clawson point.
C'
E P
C'
Cl K A J
B
B'
C I
L
L
A'
N
N
M
Fig. 1 The extangential triangle
1. Introduction In the diagram we see a triangle ABC and its incircle and its three ex-circles. The external common tangents to the three pairs of excircles (other than the sides of ABC) form a triangle 1
A'B'C'. It is shown that ABC and A'B'C' are in perspective, with perspector the orthocentre of the intouch (or contact) triangle. This is the triangle consisting of the three points on the sides of ABC where the incircle touches them. The Desargues axis of perspective LMN is shown, and points L, M, N have the property that lines joining pairs of centres of the ex-circles pass through them. We also show that the triangles A'B'C' and the orthic triangle are homothetic with vertex of enlargement the Clawson point. Work is carried out using areal co-ordinates with ABC as triangle of reference. 2. The ex-circles and their points of contact with the sides of ABC It is well known that the ex-circle opposite A has equation (a + b + c)2x2 + (a + b – c)2y2 + (c + a – b)2z2 – 2(a + b – c)(c + a – b)yz + 2(a + b + c)(c + a – b)zx + 2(a + b + c)(a + b – c)xy = 0. (2.1) The ex-circles opposite B and C have similar equations that may be written down from Equation (2.1) by cyclic change of x, y, z and a, b, c. The centres I, J, K of the ex-circles have co-ordinates I(– a, b, c), J(a, – b, c) and K(a, b, – c) so the line JK has equation cy + bz = 0. (2.2) This meets BC at the point L with co-ordinates L(0, b, – c). Similarly M = CA^KI and N = AB^IJ have co-ordinates M(– a, 0, c) and N(a, – b, 0). L, M, N are collinear on the line with equation x/a + y/b + z/c = 0, (2.3) which is the Desargues’ axis of perspective of triangles ABC and IJK. We now tabulate, without details, the points of contact L1, L2, L3 of the ex-circle opposite A with the sides BC, CA, AB respectively. These are L1(0, c + a – b, a + b – c), L2(b – c – a, 0, a + b + c) and L3(c – a – b, a + b + c, 0). Similarly the points of contact M2, M3, M1 of the ex-circle opposite B with the sides CA, AB, BC respectively are M2(b + c – a, 0, a + b – c), M3(a + b + c, c – a – b, 0), M1(0,, a – b – c, a + b + c). And again the points of contact of the ex-circle opposite C with the sides AB, BC, CA respectively N3(b + c – a, c + a – b, 0), N1(0, a + b + c, a – b – c), N2(a + b + c, 0, b – c – a). 3. The extangential triangle With L as defined in section 1 we assert that the tangent from L to the ex-circle opposite B also touches the ex-circle opposite C. The proof is as follows:
2
Any line through L has an equation of the form x = k(cy + bz) for varying k. The value of k for this line to touch the excircle opposite B is k = – (b + c)/(bc) and the co-ordinates of the point of contact J1 is (x, y, z), where x = (a + b – c)(b + c)2, (3.1) 2 y = – b (a + b + c), (3.2) 2 z = c (b + c – a). (3.3) The line LJ1 may now be seen to touch the ex-circle opposite C at the point K1 with co-ordinates (x, y, z), where x = (c + a – b)(b + c)2, (3.4) 2 y = b (b + c – a), (3.5) 2 z = – c (a + b + c). (3.6) The line LJ1K1 is now the side B'C' of the extangential triangle, with equation bcx + (b + c)(cy + bz) = 0. Similarly the sides C'A' and A'B' have equations cay + (c + a)(az + cz) = 0, and abz + (a + b)(bx + ay) = 0. The lines with equations (3.8) and (3.9) meet at A' with co-ordinates (x, y, z), where x = – a2(a + b + c), y = b(c + a)(a + b – c), z = c(a + b)(c + a – b).
(3.7)
(3.8) (3.9)
(3.10) (3.11) (3.12)
The co-ordinates of B' and C' may be written down from equations (3.10) – (3.12) by cyclic change of x, y, z and a, b, c. 4. ABC and A'B'C' are in perspective It may now be shown that AA', BB', CC' are concurrent at the point P with co-ordinates (x, y, z), where x = a(b + c)/(b + c – a), (4.1) y = b(c + a)/(c + a – b), (4.2) z = c(a + b)/(a + b – c). (4.3) P is the orthocentre of the intouch triangle, X65 in Kimberling’s Encyclopaedia of Triangle Centres. It is interesting to note that line LMN is also the Desargue’s axis of perspective for these two triangles. 3
5. The orthic triangle of triangle ABC and triangle A'B'C' are homothetic The third side of the orthic triangle has equation (b2 + c2 – a2)x = (c2 + a2 – b2)y + (a2 + b2 – c2)z.
(5.1)
The line B'C' with equation (3.7) meets this line at the point (b2 – c2, – b2, c2), which is on the line at infinity. It follows that these two lines are parallel. This is also the case for the other two pairs of sides and hence the triangles are homothetic. The lines through corresponding pairs of vertices are concurrent at a point Cl with co-ordinates (x, y, z), where x = a/(b2 + c2 – a2), y = b/(c2 + a2 – b2), z = c/(a2 + b2 – c2). (5.2) The point Cl is called the Clawson point, X19 in Kimberling’s Encyclopaedia of Triangle Centres. Although John Clawson (1) studied this point in 1925, it was studied earlier by Lemoine. Reference 1. Kimberling, C. "Encyclopedia of Triangle Centers: X(19) = Clawson Point." http://faculty.evansville.edu/ck6/encyclopedia/ETC.html#X19.
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Article: CJB/2011/126
Some Properties of the Ex-Circle configuration Christopher Bradley Abstract: The ex-circle configuration is studied and it is shown in particular how the midpoints of the sides of the ex-central triangle lead to several circles and perspectives, as well as a new (to me) triangle with vertices on the sides of the original triangle and two new (to me) points on the circumcircle.
N2
Ic
bc
M3 A
C' Ib M2
N3 P ca
B'
O Q
A' B N1
L1
L ab
L3
M1
C
Y S
T Z
N
L2 Ia
M
Fig. 1 The ex-circles and the ex-central triangle
1
1. Introduction Triangle ABC is given together with its ex-circles, centres Ia, Ib, Ic. The midpoints of the sides of triangle IaIbIc are denoted by bc, ca, ab. The points where the ex-circle opposite A touches the sides BC, CA, AB are L1, L2, L3 respectively. The points where the ex-circle opposite B touches the sides BC, CA, AB are M1, M2, M3 respectively. The points where the ex-circle opposite C touches the sides BC, CA, AB are N1, N2, N3 respectively. The points that are internal to the sides are L1, M2, N3 and form the ex-touch triangle. However, this triangle does not feature in what follows. It is shown that the six points where the ex-circles touch the sides of ABC externally lie on a conic. The circle M2, N2, M3 also passes through N3 and has centre bc. Similarly the circle N3L3N1L1 has centre ca and the circle L1M1L2M2 has centre ab. These last two circles meet at L1 and again at a point A'. Points B' and C' are similarly defined. The circle AM3N2 also passes through bc, the circle BN1L3 passes through ca and the circle CL2M1 passes through ab. We now have three triangles ABC, IaIbIc and bc ca ab and a fourth triangle LMN with its vertices L, M, N on BC, CA, AB respectively is defined by the relations L = BC^ ca ab, M = CA^ ab bc and N = AB^ bc ca. Lines Ia bc, Ib ca, Ic ab are concurrent at a point P and AL1, BM2, CN3 are concurrent at a point Q. Point Q is identified as Nagel’s point and P is identified as X165, the centroid of the ex-central triangle. It is now shown (i) that AA', BB', CC' are concurrent at O, the circumcentre of ABC , (ii) that MN, NL, LM pass respectively through Ia, Ib, Ic, (iii) that AL, BM, CN are concurrent at S, (iv) that S lies on the circumcircle of ABC, (v) that L bc, M ca, N ab are concurrent at T, (vi) that T also lies on the circumcircle of ABC, (vii) the Desagues’ axis of perspective XYZ of triangles LMN and IaIbIc coincides with the line ST, with LIa, MIb, NIc being parallel. In the sections that follow the above results are established using areal co-ordinates with ABC as triangle of reference. 2. The ex-circles and their points of contact with the sides of ABC and the point Q For convenience we tabulate elementary results. The equation of the ex-circle opposite A is (a + b + c)2x2 + (a + b – c)2y2 + (c + a – b)2z2 – 2(a + b – c)(c + a – b)yz + 2((a + b + c)(c + a – b)zx + 2(a + b + c)(a + b – c)xy = 0. (2.1) The equations of the ex-circles opposite B and C may be written down from Equation (2.1) by cyclic change of x, y, z and a, b, c.
2
We now give, without further ado, the co-ordinates of the nine points of contact, with labels as described in Section 1. L1: (0, c + a – b, a + b – c), L2: (b – c – a, 0, a + b + c), L3: (c – a – b, a + b + c, 0). M2: (b + c – a, 0, a + b – c), M3:(a + b + c, c – a – b, 0), M1:(0, a – b – c, a + b + c). N3: (b + c – a, c + a – b, 0), N1: (0, a + b + c, a – b – c), N2:(a + b + c, 0, b – c – a). The point Q is the point of concurrence of AL1, BM2, CN3. It has co-ordinates (b + c – a, c + a – b, a + b – c) and may therefore be identified as Nagel’s point.
3. The conic through L2, L3, M3, M1, N1, N2 This is easily obtained with a computer algebra package and has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where u = v = w = (a + b + c)(b + c – a)(c + a – b)(a + b – c), f = (a2 + (b + c)2)(a + b – c)(c + a – b), g = (b2 + (c + a)2)(b + c – a)(a + b – c), h = (c2 + (a + b)2)(c + a – b)(b + c – a). (3.2) 4. The circles Σmn = M2M3N2N3 and Σnl = N3N1L3L1 and Σlm = L1L2M1M2 The circle Σmn passes through the four points M2, M3, N2, N3 and has an equation of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0, (4.1) where u = ¼ (a + b – c)(c + a – b), v = ¼ (b + c – a)(a + b – c), w = ¼ (c + a – b)(b + c – a). (4.2) Circles Σnl and Σlm have equations that may be written down from that of Σmn by cyclic change of x, y, z and a, b, c. The centres of these three circles are the points bc, ca, ab respectively. These points are the midpoints of IbIc, IcIa, IaIb respectively. The co-ordinates of bc are (a2, b(c – b), c(b – c)) and those of ca, ab may be written down by cyclic change of x, y, z and a, b, c. Circles Σmn and Σnl meet at N3 and a point we label C'. The co-ordinates of C' may be obtained because one point of intersection is already known. Its co-ordinates are (x, y, z), where x = a2(b2 + c2 – a2), y = b2(c2 + a2 – b2), z = (a + b)(a3 + b3 – a2b – ab2 – ac2 – bc2 + 2abc). (4.3) Points A' and B' are defined similarly and it is immediately apparent that AA', BB', CC' are concurrent at the circumcentre O of triangle ABC. 3
5. The point P Since bc, ca and ab are the midpoints of the sides of the ex-central triangle, it is obvious that Ia bc, Ib ca, and Ic ab are concurrent at a point that is the centroid of the ex-central triangle. Its xco-ordinate is x = a( – 3a2 + 2a(b + c) + (b – c)2). (5.1) Its y- and z- co-ordinates follow from Equation (5.1) by cyclic change of a, b, c. P is identified as X165 in Kimberlings’ Encyclopaedia of Triangle Centres and is, of course, the centroid of the excentral triangle. 6. The triangle LMN The equation of the line ca ab is bcx + c(a – c)y + b(a – b) = 0.
(6.1)
This meets the side BC at the point L with co-ordinates L(0, b(a – b), c(c – a)). Similarly points M and N have co-ordinates M(a(a – b), 0, c(b – c)) and N(a(c – a), b(b – c), 0). The line MN has equation bc(b – c)x + ca(a – c)y + ab(b – a)z = 0.
(6.2)
It is now easily checked that MN passes through Ia(– a, b, c). Similarly NL passes through Ib and LM passes through Ic. 7. The points S and T The equation of the line AL is c(a – c)y + b(a – b)z = 0,
(7.1)
and the equation of the line BM is c(b – c)x + a(b – a)z = 0.
(7.2)
These meet at the point S with co-ordinates (a(a – b)(a – c), b(b – c)(b – a), c(c – a)(c – b)). It may be checked that S also lies on CN and remarkably on the circumcircle of triangle ABC. The equation of bc L is b2c(b – c)x + ca(c – a)2y + b2a(b – a)z = 0. 4
(7.4)
This line and the lines ca M and ab N are concurrent at the point T with co-ordinates (a2(a – b)(a – c), b2(b – c)(b – a), c2(c – a)(c – b)). Again it may be checked that T also lies on the circumcircle of triangle ABC. The equation of ST is bc(b – c)2x + ca(c – a)2y + ab(a – b)2z = 0.
(7.5)
The line ST, as we soon see, is of some significance in the configuration. 8. The points X, Y, Z We define the points X, Y, Z to be X = MN^IbIc, Y = NL^IcIa, Z = LM^IaIb. MN has Equation (6.2) and IbIc has equation cy + bz = 0. From which X has co-ordinates X(a(2a – b – c), b(c – b), c(b – c)). It may now be checked that X lies on the line ST. So do Y and Z. So the Desargues; axis of perspective of triangles XYZ and IaIbIc is the line ST. As we show finally the perspector is a point on the line at infinity with co-ordinates (a(b – c), b(c – a), c(a – b)) showing that the lines LIa, MIb, NIc are parallel. 9. The circles A bc N2 M3, B ca L3 N1, C ab M1 L2 The four points A, bc, N2, M3 may be shown to lie on the circle with equation of the form (4.1) with u = 0, v = – ½ c(a + b + c), w = – ½ b(a + b + c). Equations of the other two circles may be found from this by cyclic change of x, y, z and a, b, c. The ex-circle configuration thus has seven circles and many more interesting points, lines and triangles than appear at first sight.
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Article: CJB/2011/127
When a Circle passes through a Vertex and cuts the Opposite Side Christopher Bradley Abstract: A surprising configuration emerges when a circle passes through a vertex and cuts the opposite side. Two pairs of touching circles and two circles each passing through seven key points emerge and additional points on the original circle are also created. It seems unlikely that a synthetic solution will exist, as some of the co-ordinate working is technically very difficult.
A N
O
Q' E F
M
K'
P'
L' C
B
D'
D K L P
Q
Fig. 1 Seven inter-related circles 1
1. Introduction The construction of Fig. 1 is delightfully straightforward. What is given initially is triangle ABC and a circle Σ passing through A that cuts CA at E, AB at F and BC in real distinct points D and D'. Circle BDF centre K, circle BD'F centre L, circle CDE centre K' and circle CD'E centre L' are now drawn. We prove that circles BDF and CDE touch at D and that circles BD'F and CD'E touch at D'. It then emerges that the common tangents meet at a point N on Σ. We then prove that lines KK' and LL' meet at a point M also lying on Σ, and furthermore that NM is a diameter of Σ. The centre of Σ is denoted by O. Line OD meets circle BDF at P and circle CDE at Q’. Also line OD' meets circle BD'F at Q and circle CD'E at P'. We prove finally that points O, F, K, L, P, Q, M are concyclic and that points O, Q', P', K', L', M, E are concyclic. We use areal co-ordinates with ABC as triangle of reference. 2. Circle AED'DF In areal co-ordinates a circle has an equation of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0,
(2.1)
where u, v, w are constants depending on the co-ordinates of the three points that define the circle. I setting up an account of the present configuration it is important to start with co-ordinates of D and D', rather than E and F, otherwise square roots appear rendering the account impossible. So we choose D to have co-ordinates D(0, p, 1 – p) and D' to have co-ordinates D'(0, 1 – q, q). Then the circle ADD' has an equation of the form (2.1), with u = 0, v = a2q(p – 1), w = a2p(q – 1). (2.2) It is now a straightforward matter to find the co-ordinates of E and F, which are E(a2p(1 – q), 0, a2p(q – 1) + b2) and F(a2q(1 – p), a2q(p – 1) + c2, 0). 3. Circles BDF, BD'F, CDE, CD'E Using the co-ordinates of the points concerned we now work out and present the values of u, v, w for each of these circles. BFD: BFD': CED:
u = a2q(1 – p) – c2, v = 0, w = – a2p. u = a2q(1 – p) – c2, v = 0, w = a2(q – 1). u = a2p(1 – q) – b2, v = a2( p – 1), w = 0. 2
(3.1) (3.2) (3.3)
CED':
u = a2p(1 – q) – b2, v = – a2q, w = 0.
(3.4)
4. The common tangents The equation of the tangent at D to the circle BDF is (a2(p – q) – b2 + c2)x + a2(p – 1)y + a2pz = 0.
(4.1)
It may now be checked that this line touches circle CED. It also meets circle Σ at a point N with co-ordinates N(a2, a2(p – q) – b2 + c2, – (a2(p – q) – b2 + c2)). (4.2) The line ND' has equation (a2(p – q) – b2 + c2)x = a2qy + a2(q – 1)z.
(4.3)
It may now be checked that ND' touches both circles BFD' and CED' at D'. 5. Circle centres For circles with an equation of the form (2.1) the centre has co-ordinates (x, y, z), where x = a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w), (5.1) 4 2 2 2 2 2 y = b – b (c + a + 2v – w – u) + (c – a )(w – u), (5.2) 4 2 2 2 2 2 z = c – c (a + b + 2w – u – v) + (a – b )(u – v). (5.3) Values of u, v, w for each of the circles are given in Sections 2 and 3. Using these values and Equations (5.1) – (5.3) we find the centres to be Circle AED'DF, centre O: x = a2(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1)), y = a4p(1 – q) – a2(b2(p(q + 1) – 2q + 1) + c2p(1 – q)) + b4 – b2c2, z = a4q(1 – p) + a2(b2q(p – 1) – c2(p(q – 2) + q + 1)) – b2c2 + c4. Circle BDF, centre K: x = a2(a2(p(2q – 1) – 2q + 1) + b2(p – 1) +c2(1 – p)), y = a4(q – p(q – 1)) + a2(c2(p(q – 1) – q – 1)) – b2(p(q + 1) – q + 1)) + (b2 – c2)2, z = a4q(1 – p) + a2(b2q(p – 1) – c2(p(q – 2) – q + 2)). Circle BD'F, centre L: x = a2q(a2(2p – 1) – b2 + c2), y = a4(1 – pq) – a2(b2(pq – 2q + 2) – c2(2 – pq)) + (b2 – c2)2, 3
(5.4) (5.5) (5.6)
(5.7) (5.8) (5.9)
(5.10) (5.11)
z = a4q(1 – p) + a2(b2q(p – 1) – c2(pq + q)). Circle CDE, centre K': x = a2p(a2(2q – 1) + b2 – c2), y = a4p(1 – q) – a2(b2p(q + 1) + c2p(1 – q)), z = a4(1 – pq) + a2(b2(pq – 2) – c2(p(q – 2) + 2)) + (b2 – c2)2. Circle CD'E, centre L': x = a2(a2(2p(q – 1) – q + 1) + b2(1 – q) + c2(q – 1)), y = a4p(1 – q) – a2(b2(p(q – 1) – 2q + 2) + c2p(1 – q)), z = a4(q – p(q – 1)) + a2(b2(p(q – 1) – q – 1) – c2(p(q – 1) + q + 1)) + (b2 – c2)2.
(5.12)
(5.13) (5.14) (5.15)
(5.16) (5.17) (5.18)
6. The point M and two additional lines passing through M The work has already become technically complicated, but now becomes even more so. The coordinates of N are given in (4.2) and those of O in (5.4) - (5.6). It is therefore possible to obtain the equation of NO. However, since this only serves to find the co-ordinates of M, we can instead use the equation M = 2O – N for each co-ordinate in turn, since O is the midpoint of NM. The result is the co-ordinates of M, which are (x, y, z), where x = – a2(a4(2q – 1)(2p – 1) + 2a2(b2 – c2)(p – q) – (b2 – c2)2), (6.1) 2 2 2 4 2 2 2 2 2 2 y = (a + b – c )(a (p(2q – 1) + a (b (p – q + 1) – c (p – q – 1)) – (b – c ) ), (6.2) z = (c2 + a2 – b2)(a4(p(2q – 1) – q) + a2(b2(p – q + 1) – c2(p – q – 1)) – (b2 – c2)2) (6.3) It may now be verified by determinantal methods using the co-ordinates of the points involved that lines KDK' and LD'L' pass through M. 7. The lines OD, OD', and the points P, P', Q, Q' The equation of OD is (a4p(p – 1)(2q – 1) + a2(b2(p – 1)(p – 2q + 1) + c2p(2q – p)) + (c2 – b2)(b2(p – 1) – c2p))x + a2(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1))(p – 1)y + pz) = 0. (7.1) Line OD meets circle BDF at D and a point P with co-ordinates (x, y, z), where x = a2(1 – p)(a2(2p – 1) – b2 + c2)(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1)), (7.2) 6 4 2 3 2 2 2 y = a p(p – 1)(p(2q – 1) – q) + a (b (2p q + 4p (1 – 2q) + p(8q – 5) – 2q + 1) – c p(2p q – 4pq + 2q + 1)) + a2(b4(p3 – p2(q + 5) + p(3q + 4) – 2q) – 2b2c2(p3 – p2(q + 3) + p(2q + 1) – q) + c4p(p2 – p(q + 1) + q + 2)) + b6(p – 1) + b4c2(2 – 3p) + b2c4(3p – 1) – c6p, (7.3) 4
z = a2(p – 1)(a2(p – 1) + b2(1 – p) + c2(p + 1))(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1)).
(7.4)
The equation of OD' is (a4q(q – 1)(2p – 1) + a2(b2q(2p – q) + c2(1 – q)(2p – q – 1)) + (c2 – b2)(b2q + c2(1 – q))) + a2(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1))(qy + (q – 1)z) = 0. (7.5) Line OD' meets circle BD'F at D' and a point Q with co-ordinates (x, y, z), where x = – a2q(a2(2q – 1) + b2 – c2)(a2(p(2q – 1) – q + 1) + b2(p – q – 1) – c2(p – q + 1)), (7.6) 6 4 2 2 2 2 2 2 y = a q(q – 1)(p(2q – 1) – q) + a (b q(2p(q + q – 1) – 2q + 2q – 1) + c (1 – q)(2pq – 2q – 1)) + a2(b4q(p(q + 1) – q2 – 3q + 2) – 2b2c2(pq2 – q3 – q2 + 2q – 1) + c4(q – 1)(pq – q2 – 2)) – b6q + c2(b4(3q – 1) + b2c2(2 – 3q) + c4(q – 1)). (7.7) Points P', Q' may be determined in similar fashion, but we do not need their co-ordinates. 8. Circles OFKLPQM and OQ'P'K'L'ME Circle OFKLPQM turns out to have equation of the form (2.1) with u = (1/t)( a6q(p2(2q – 1) + p(1 – 3q) + q) + a4(b2q(p2 – pq + q – 1) – c2(p2q – p(q2 + 4q – 1) + q2 + 2q)) – a2(b4q(p – 1) – b2c2(p(2q + 1) – 3q + 1) + c4(p(q + 1) – 2q – 1)) – c2(b2 – c2)2) , (8.1) 4 2 2 2 v = (1/t) (a q(p – 1)(a (p(2q – 1) – q + 1) + b (p – q – 1) – c (p – q + 1))), (8.2) 2 4 2 2 2 2 2 2 2 w = (1/t)(a (a (pq – q + 1)(p(2q – 1) – q) + a (b (p q – p(q + 3q – 2) + q ) – c (p q – pq(q + 1) + (q2 – 1))) + b4(q – p) + b2c2(2p – 2q + 1) – c4(p – q + 1))). (8.3) Here t = a4 + b4 + c4 – 2b2c2 – 2c2a2 – 2a2b2.
(8.4)
The symmetrical nature of the configuration ensures that the points O, Q', P', K', L', M, E also lie on a circle. 9. Sets of parallel lines AN and BC: These meet at (0, 1, – 1) on the line at infinity and are therefore parallel. BKQ and CK'P': These meet at (a2(1 – 2q) – b2 + c2, a2(q – 1) + b2(q + 1) + c2(1 – q), a2q – b2q + c2(q – 2)) on the line at infinity and are therefore parallel. BLP and CL'Q': these meet at (a2(1 – 2p) + b2 – c2, a2p + b2(p – 2) – c2p, a2(p – 1) + b2(1 – p) + c2(p + 1)) on the line at infinity and are therefore parallel. 5
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Article: CJB/2011/128
The Symmetry of a Scalene Triangle Christopher Bradley Abstract: Points dividing each side of a scalene triangle in a fixed ratio leads to a configuration with two sets of three equal circles and a conic. Many results are self-evident but the equations of the circles and conic are recorded.
A
Ca W
C'
V Sa
N
B'
O
M G
Sb
Sc
Cb Cc
B
L
U
C
A'
Fig. 1 Two sets of three equal circles and a conic drawn in a scalene triangle 1. Introduction 1
Fig. 1 was constructed as follows: A point U was selected on BC nearer to B than C. The line through U parallel to CA meets AB at N and the line through U parallel to AB meets CA at V. The line through N parallel to BC meets CA at M and the line through V parallel to BC meets AB at W. It is then shown that the six points U, V, W, L, M, N lie on a conic. Circles ANM, BWL and CUV are drawn and because of the symmetrical distribution of points obviously have equal radius. Similarly, so do circles AWV, BNU, and CLM. Centres of circles are shown as well as three points A' = NU^ML, B' = WV^LM, C' = VW^UN. The points A, B, C, A', B', C' do not lie on a conic. 2. The six points and the conic If U has co-ordinates U(0, p, 1 – p), 1 > p > ½, then it is easily shown that the co-ordinates of the other five key points are L(0, 1 – p, p), W(p, 1 – p, 0), N(1 – p, p, 0), V(p, 0, 1 – p),M(1 – p, 0, p). The conic through these six points has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(2.1)
where u = v = w = 2p(p – 1) and f = g = h = 2p2 – 2p + 1.
(2.2)
3. The six circles Circles have the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(3.1)
u = 0, v = c2(p – 1), w = b2(p – 1).
(3.2)
u = 0, v = – c2p, w = – b2p.
(3.3)
Circle AMN has
Circle AWV has
The other four circles may be obtained by cyclic change o u, v, w and a, b, c.
Flat 4, Terrill Court, 12-14, Apsley Road BRISTOL BS8 2SP.
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Article: CJB/2011/129
The Neuberg Cubic Christopher Bradley Abstract: Analysis is made of the Neuberg triangle cubic. This is the locus of points P with the property that the reflection of P in the sides of a triangle ABC leads to a triangle which is in perspective with ABC.
A W
V P
Q
B
C
U
Fig. 1 When UVW is in perspective with ABC then P lies on the Neuberg Cubic 1. Introduction In Fig. 1 P is a point and its image with respect to reflections in BC, CA, AB are the points U, V, W respectively. When AU, BV, CW are concurrent at a point Q then P lies on the Neuberg cubic. In this article the equation of the Neuberg cubic is obtained using areal co-ordinates. 1
Checks are made confirming that it passes through the circumcentre, the incentre and ex-centres, the Fermat points and the isodynamic points. 2. The points U, V, W We suppose that P has co-ordinates P(d, e, f). Then U, the image of P in BC, has co-ordinates (x, y, z), (see Bradley’s Algebra of Geometry p422), where x = – d, (2.1) 2 2 2 2 y = e + d(a + b – c )/a (2.2) 2 2 2 2 z = f + d(c + a – b )/a . (2.3) Points V, W have co-ordinates which may be written down from (2.1) – (2.3) by cyclic change of x, y, z and a, b, c and d, e, f. The equation of AU is (a2(f + d) – d(b2 – c2))y = (a2(d + e) + d(b2 – c2))z.
(2.4)
and the equations of BV, CW may be written down from Equation (2.4), again by cyclic change of x, y, z and a, b, c and d, e, f). 3. The point Q = AU^BV and the condition that CW passes through Q Lines AU and BV meet at the point Q with co-ordinates (x, y, z), where x = (a2e + b2(d + e) – c2e)(d(b2 – c2) – a2(f + d)), y = (a2e – b2(e + f) – c2e)(a2(d + e) + d(b2 – c2)), z = (a2e – b2(e + f) – c2e)(a2(f + d) – d(b2 – c2)).
(3.1) (3.2) (3.3)
The condition that Q also lies on CW may now be obtained and is (d2ec2 – f2ea2) (a4 + a2(b2 – 2c2) + (c2 – b2)(2b2 + c2)) + (e2fa2 – d2fb2)(a4 + a2(c2 – 2b2) + (b2 – c2)(b2 + 2c2)) + (e2dc2 – f2db2)(2a4 – a2(b2 + c2) – (b2 – c2)2).
(3.4)
Putting d = x, e = y, f = z in Equation (3.4) gives the locus of Q when triangles DEF and ABC are in perspective, in other words is the equation of the Neuberg cubic. Checks may now be made confirming that it passes through the circumcentre, the incentre and ex-centres, the Fermat points and the isodynamic points. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
Article: CJB/2011/130
An Extension to the Theory of Hagge Circles Christopher Bradley
A D* W*
X B*
Y
C*
O H* P
E*
Q
V*
H U* O*
Z B
C
Pg
F* A*
Fig. 1 Two related triangles, a circle with significance in both triangles 1. Introduction One starts with triangle ABC and some point Q not on the sides. Triangle ABC is then rotated about Q by 180o to form a second triangle A*B*C*. So AQ = QA* etc. The point A* is now reflected in BC to produce a point X, B* is reflected in CA to produce Y and C* is reflected in AB to produce Z. Triangle XYZ turns out to be similar to triangle ABC, and circle XYZ we prove passes through H*, the orthocentre of triangle A*B*C* and has centre O, the circumcentre 1
of ABC. This is the dual significance of circle XYZ. With respect to ABC its centre is the same as its circumcentre and with respect to A*B*C*, since it passes through H*, it is a Hagge circle. It is also the case that A, B, C, A*, B*, C* lie on a conic. It follows that A*, H*, X are collinear as are B*, H*, Y and C*, H*, Z. The following properties now follow from the fact that XYZH* is a Hagge circle. A point Pg exists such that OO*PgH* is a parallelogram. The isogonal conjugate of Pg, the point P is the point by which there is another construction of points X, Y, Z. To be precise if A*P, B*P, C*P meet the circumcircle of A*B*C* at D*, E*, F* respectively. Then if D*, E*, F* are reflected in B*C*, C*A*, A*B* respectively, the images U*, V*, W* lie on circle XYZ and are such that U*P, V*P, W*P meet the Hagge circle again at X, Y, Z respectively. Also shown in Fig. 1 is a line through P which is the axis of indirect similarity of triangles A*B*C* and XYZ, P being the centre of the similarity. In the following sections we prove, using areal co-ordinates with ABC as triangle of reference, the features additional to what is already known about Hagge circles. See Bradley’s ‘Algebra of Geometry’ p 419-420 for the co-ordinate geometry of Hagge circles. 2. Points A*, B*, C*, X and H* Suppose Q has normalised co-ordinates Q(1 – u – v, u, v). Then, since Q is the midpoint of AA*, it follows that A* has co-ordinates A*(1 – 2u – 2v, 2u, 2v). Similarly B* has co-ordinates B*(2 – 2u – 2v, 2u – 1, 2v) and C* has co-ordinates C*(2 – 2u – 2v, 2u, 2v – 1). The reflection of A* in BC is the point X with co-ordinates (x, y, z), where x = 2u + 2v – 1, y = 2u + (1 – 2u – 2v)(a2 + b2 – c2)/a2, z = 2v + (1 – 2u – 2v)(c2 + a2 – b2)/a2.
(2.1) (2.2) (2.3)
The point H has normalised co-ordinates (x, y, z), where x = (1/k)(a2 + b2 – c2)(c2 + a2 – b2), y = (1/k)(b2 + c2 – a2)(a2 + b2 – c2), z = (1/k)(c2 + a2 – b2)(b2 + c2 – a2),
(2.4) (2.5) (2.6)
where k = 2(b2c2 + c2a2 + a2b2) – a4 – b4 – c4. The point H* lies on HQ and is such that HQ = QH*. It follows that H* has co-ordinates (x, y, z), where x = 3 – 2u – 2v – (2/k)(a2(b2 + c2) – (b2 – c2)2), (2.7) 2 2 2 2 y = 2u – 1 + (2/k) b (c + a – b ), (2.8) 2 2 2 2 z = 2v – 1 + (2/k)c (a + b – c ). (2.9) It may now be proved by the determinantal method that A*, H*, X are collinear.
2
3. Points B*, C*, Y, Z and circle XYZ B*, C* are the points on BQ, CQ respectively such that BQ = QB* and CQ = QC*. Their coordinates are therefore B*(2 – 2u – 2v, 2u – 1, 2v) and C*(2 – 2u – 2v, 2u, 2v – 1). The point Y is the reflection of the point B* in CA and so has co-ordinates (x, y, z), where x = (1/b2)(a2(2u – 1) + b2(1 – 2v) + c2(1 – 2u)), (3.1) y = 1 – 2u, (3.2) 2 2 2 2 z = (1/b )(a (1 – 2u) + b (2u + 2v – 1) – c (1 – 2u)). (3.3) The point Z is the reflection of the point C* in AB and so has co-ordinates (x, y, z), where x = (1/c2)(a2(2v – 1) + b2(1 – 2v) + c2(1 – 2u)) (3.4) 2 2 2 2 y = (1/c )(a (1 – 2v) – b (1 – 2v) + c (2u + 2v – 1)), (3.5) z = 1 – 2v. (3.6) The equation of the circle XYZ is of the form a2yz + b2zx + c2xy + (lx + my + nz)(x + y + z) = 0,
(3.7)
where l, m, n depend upon the co-ordinates of X, Y, Z. After inserting the co-ordinates of X, Y, Z in Equation (3.7) and solving the three equations obtained we find l =m = n = a2(1 – 2v)(2u – 1) + (2u + 2v – 1)(b2(2v – 1) + c2(2u – 1)). (3.8) It may now be checked that H* lies on circle XYZ. This means we have a Hagge circle and the simplicity of the equation representing it, in terms of the co-ordinates of Q, surely means that this way of determining Hagge circle is superior to any other. The fact that A*, H*, X etc. are collinear shows that the points X, Y, Z correspond precisely to the points associated with Hagge’s construction. We repeat from the Introduction how that construction is performed. The following properties now follow from the fact that XYZH* is a Hagge circle. A point Pg exists such that OO*PgH* is a parallelogram. The isogonal conjugate of Pg, the point P is the point by which there is another construction of points X, Y, Z. To be precise if A*P, B*P, C*P meet the circumcircle of A*B*C* at D*, E*, F* respectively. Then if D*, E*, F* are reflected in B*C*, C*A*, A*B* respectively, the images U*, V*, W* lie on circle XYZ and are such that U*P, V*P, W*P meet the Hagge circle again at X, Y, Z respectively. Also shown in Fig. 1 is a line through P which is the axis of indirect similarity of triangles A*B*C* and XYZ, P being the centre of the similarity 3
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Article: CJB/2011/131
Construction of Circles always having centre the Nine-Point Centre Christopher Bradley Abstract: From any general point P reflections in BC. CA, AB give D, E, F. The midpoints of AD, BE, CF are U, V, W respectively. U, V, W are shown to lie on a circle with centre the NinePoint centre.
A' A
F U C'
O W
N
E V
P C
B D
B'
Fig. 1 Illustrating the construction to produce a circle UVW centre N, the Nine-Point centre 1. Introduction
1
Given a triangle ABC, then from any general point P reflections in BC. CA, AB determine points D, E, F. The midpoints of AD, BE, CF are U, V, W respectively. In this article U, V, W are all shown to lie on a circle with centre the Nine-Point centre N. This is suggested by the fact that if P lies at O, the circumcentre of ABC, then U, V, W coincide with N, providing a point circle. It is also the case that triangle UVW is indirectly similar to triangle ABC. We use areal coordinates with ABC the triangle of reference to prove these assertions. 2. The points D, E, F and U, V, W We take P to have co-ordinates P(d, e, f). Then the normalized co-ordinates of D, the reflection of P in BC are (x, y, z), where x = – d, y = e + d(a2 + b2 – c2)/a2, z = f + d(c2 + a2 – b2). (2.1) The normalized co-ordinates of E, F may be written down from Equation (2.1) by cyclic change of x, y, z and a, b, c, and d, e, f. The midpoint of AD is the point U, which has normalized co-ordinates (x, y, z), where x = (1 – d)/2, y = (a2(d + e) + d(b2 – c2))/(2a2), z = (a2(d + f) – d(b2 – c2))/(2a2).
(2.2) (2.3) (2.4)
The normalized co-ordinates of V and W may be written down from Equations (2.2) – (2.4) by cyclic change of x, y, z and a, b, c, and d, e, f. 3. The circle UVW All circles in areal co-ordinates have the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(3.1)
where u, v, w depend on the co-ordinates of three points lying on the circle. Inserting the coordinates of U, V, W and solving for u, v, w and simplifying by putting f = 1 – e – d we obtain u = (1/4)(a2(de + e2 – e + 1) + b2(d2 + d(e – 1) – 1) – c2(1 + de)), (3.2) 2 2 2 2 2 v = (1/4)(a (de + e – e – 1) + b (d + d(e – 1) + 1) – c (1 + de)), (3.3) 2 2 2 2 2 w = (1/4)(a (de + e – e – 1) + b (d + d(e – 1) – 1) + c (1 – de)). (3.4) Using standard formulae for the centre of a conic (see Bradley, The Algebra of Geometry p. 57) we obtain the co-ordinates of N, which are proportional to (x, y, z), where x = a2(b2 + c2) – (b2 – c2)2, (3.5) 2 2 2 2 2 2 y = b (c + a ) – (c – a ) , (3.6) 2 2 2 2 2 2 z = c (a + b ) – (a – b ) . (3.7) 2
4. Triangle XYZ is indirectly similar to triangle ABC This is illustrated in Fig. 1, where the line which is the perpendicular bisector of ON is shown, where O is the circumcentre of triangle ABC. Reflection of ABC in this line produces the triangle A'B'C' and this is an enlargement of triangle XYZ centre N and scale factor BC/YZ. This is, of course, only a pictorial verification of the indirect similarity, but the accuracy of CABRI II plus is such that one definitely expects the result to hold. We take the co-ordinates of V and W and work out VW2/a2 = VW2/BC2. For those less familiar with areal co-ordinates if the displacement VW = (x, y, z), where x + y + z = 0 then VW2 = – a2yz – b2zx – c2xy (4.1). We actually find VW2/BC2 = (1/4a2b2c2)(a6ef – a4(b2f(3e + f – 1) + c2e(e + 3f – 1)) + a2(b4f(3e + 2(f – 1)) + b2c2(2e2 + 2e(f – 1) + 2f2 – 2f + 1) + c4e(2e + 3f – 2)) + + (c2 – b2(e + f – 1)(b4f + b2c2(e – f) – c4e)) (4.2) If in Equation (4.2) we change the right hand side by cyclic change of a, b, c and d, e, f and in the result we then put d = 1 – e – f we find it is invariant. In other words VW/BC = WU/CA = UV/AB and hence the indirect similarity is valid.
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Article: CJB/2011/132
The Cevian Conic Christopher Bradley Abstract: If P is a Cevian point in a triangle ABC and D, E, F are the feet of the Cevians, then three circles PEF, PFD, PDE meet the sides again in six other points. It is shown that these six points lie on a conic, which we call the Cevian Conic.
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Fig. 1 The Cevian Conic 1. Introduction
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Let ABC be a triangle and P a Cevian point with D, E, F the feet of the Cevians on BC, CA, AB respectively. Circle EFP meets CA again at L and AB at U, circle FDP meets AB again at M and BC at V and circle DEP meets BC again at N and CA at W. In this article we show points L, M, N, U, V, W lie on a conic, which we call the Cevian Conic of P. This conic is a circle when P is Gergonne’s point. Proof is given using areal co-ordinates with ABC as triangle of reference. 2. Points P, D, E, F and circle EFP Suppose P has co-ordinates P(d, e, f). The co-ordinates of D, E, F are D(0, e, f), E(d, 0, f), F(d, e, 0). Circle EFP has equation of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0. (2.1) Inserting the co-ordinates of P, E, F we find that circle EFP has u = ef(a2(d + e)(d + f) – d(b2(d + e) + c2(d + f)))/{d(d + e)(d + f)(d + e + f)}, v = d(b2f – c2(d + f)) – a2f(d + f)/{(d + e + f)(d + f)}, w = – (a2e(d + e) + d(b2(d + e) – c2e))/{(d + e)(d + e + f)}.
(2.2) (2.3) (2.4)
3. Points L, M, N, U, V, W This circle meets CA at L with co-ordinates (x, y, z), where x = (d + f)(a2e(d + e) + d(b2(d + e) – c2e)), y = 0, z = e(d(b2(d + e) + c2(d + f)) – a2(d + e)(d + f)). (3.1) Circle EFP also meets AB at U with co-ordinates (x, y, z), where x = (d + e)(d(b2f – c2(d + f)) – a2f(d + f)), y = f(a2(d + e)(d + f) – d(b2(d + e) + c2(d + f))), z = 0. (3.2) The co-ordinates of M and N follow from Equation (3.1) by cyclic change of x, y, z and a, b, c and d, e, f. And the co-ordinates of V and W follow from Equation (3.2) by the same process. 4. The Cevian Conic LMNUVW It may now be shown that the six points L, M, N, U, V, W all lie on the conic with equation ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, (4.1) where u = 2def(d + e)(d + f)(e + f)(a2ef – (e + f)(b2f + c2e))(a2(d + e)(d + f) – d(b2(d + e) + c2(d + f))). ` (4.2) v and w follow from Equation (4.2) by cyclic change of a, b, c and d, e, f.
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l = – ef(d + e)(d + f)(a4ef(d + e)(d + f)(d2 + (e + f)2) – a2d(d – e – f)(e + f)(b2f(d + e)(d – e + f) + c2e(d + e – f)(d + f)) – 2d2(e + f)2(b4f(d + e) – b2c2(d2 + d(e + f) + 2ef) + c4e(d + f))). (4.3) m and n follow from Equation (4.3) by cyclic change of a, b, c and d, e, f. 5. When P is Gergonne’s point What happens when P lies at Gergonne’s point is that the Cevian Conic becomes a circle, centre I, the incentre. Circle DEF is now the incircle. I is also the Miquel point for two sets of three triangles. Working is left to the reader, but the results are illustrated in Fig. 2 below.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/133
Conics in a Semi-Regular Hexagon Christopher Bradley Abstract: Hexagon AFBDCE with ABC an equilateral triangle and AD, BE, CF concurrent at the centroid P of ABC is inscribed in a conic, and as such is defined as a semi-regular hexagon. It is proved that the six circumcentres of triangles AFP, FBP, BDP, DCP, CPE, EPA are co-conic as are their six orthocentres. It is also proved that the intersections of contiguous Euler lines are co-conic. Cabri indicates that the six in-centres and six nine-point centres are also co-conic.
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Fig. 1 A semi-regular hexagon and its conics 1
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1. The co-ordinates of D, E, F We use areal co-ordinates with equilateral triangle ABC as triangle of reference. This means we may set a = b = c = 1 in known formulae. We suppose the conic in which the hexagon AFBDCE is inscribed has equation fyz + gzx + hxy = 0. (1.1) The centroid P has co-ordinates (1, 1, 1) so the equation of AP is y = z. Line AP meets the conic again at D with co-ordinates D (– f, g + h, g + h). Similarly E and F have co-ordinates E (h + f, – g, h + f) and F(f + g, f + g, – h). 2. The circumcircles of the six triangles AFP, FBP, BDP, DCP, CEP, EAP Circles with a = b = c = 1 have equations of the form yz + zx + xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
With known co-ordinates for their vertices it is straightforward to calculate u, v, w for each of the six circles. These are: AFP: u = 0, v = – (f + g – h)/(2f + 2g – h), w = – (f + g)/(2f + 2g – h). (2.2) FBP: u = – (f + g – h)/(2f + 2g – h), v = 0, w = – (f + g)/(2f + 2g – h). (2.3) BDP: u = – (g + h)/(2g + 2h – f), v = 0, w = – (g + h – f)/(2g + 2h – f). (2.4) DCP: u = – (g + h)/(2g + 2h – f), v = – (g + h – f)/(2g + 2h – f), w = 0. (2.5) CEP: u = – (h + f – g)/(2h + 2f – g), v = – (h + f)/(2h + 2f – g), w = 0. (2.6) EAP: u = 0, v = – (h + f)/(2h + 2f – g), w = – (h + f – g)/(2h + 2f – g). (2.7) 3. Centres O1 – O6 of the six circles With a = b = c = 1, the centre of a circle with equation of the form (2.1) has co-ordinates (x, y, z), where x = 2u – v – w + 1, y = 2v – w – u + 1, z = 2w – u – v + 1. (3.1) So for the six circles in Section 2 we have: AFP: x = 2, y = (f + g + h)/(2f + 2g – h), z = (f + g – 2h)/(2f + 2g – h). FBP: x = (f + g + h))/(2f + 2g – h), y = 2, z = (f + g – 2h)/(2f + 2g – h). BDP: x = (g + h – 2f)/(2g + 2h – f), y = 2, z = (f + g + h)/(2g + 2h – f). DCP: x = (g + h – 2f)/(2g + 2h – f), (f + g + h)/(2g + 2h – f), z = 2. CEP: x = (f + g + h)/(2h + 2f – g), y = (h + f – 2g)/(2h + 2f – g), z = 2. EAP: x = 2, y = (h + f – 2g)/(2h + 2f – g), z = (f + g + h)/(2h + 2f – g).
2
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
4. The Conic of Circumcentres The equation of a conic is of the form ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0.
(4.1)
From the co-ordinates in Section 3 we obtain the equation of the conic passing through O1 – O6. Its values are u = 2(5f3 – 6f2g – 6f2h – 9fg2 – 9fgh – 9fh2 + 2g3 – 3g2h – 3gh2 + 2h3), (4.2) with v and w to be obtained from Equation (4.2) by cyclic change of f, g, h. And l = 4f3 + 24f2g + 24f2h + 9fg2 + 18fgh + 9fh2 – 11g3 + 12g2h + 12gh2 – 11h3,
(4.3)
with m and n to be obtained from Equation (4.3) by cyclic change of f, g, h. 5. Modification of a known formula When a = b = c = 1 then the line perpendicular to px + qy + rz = 0 through the point with coordinates (l, m, n) is The (m(p + q – 2r) – n(r + p – 2q))x + (n(q + r – 2p) – l(p + q – 2r)y + (l(r + p – 2q) – m(q + r – 2p)z = 0. (5.1) 6. The orthocentres H1– H6 of the six triangles Equation (5.1) must be used repeatedly. The equation of AF is hy + (f + g)z = 0. The equation of the line through P perpendicular to AF is (f + g – h)x – y(f + g) + hz = 0.
(6.1)
(6.2)
The equation of the line AP is y – z = 0.
(6.3)
The equation of the line through F perpendicular to AP is (f + g – h)x – (f + g)y – (f + g)z = 0.
(6.4)
Solving equations (6.2) and (6.4) we find the co-ordinates of H1 are H1((f + g), (f + g – h), 0).
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The equation of BF is hx + (f + g)z = 0. The equation of the line through P perpendicular to BF is (f + g)x – (f + g – h)y – hz = 0.
(6.5)
(6.6)
The equation of BP is x – z = 0.
(6.7)
The equation of the line through F perpendicular to BP is (f + g)x – (f + g – h)y + (f + g)z = 0.
(6.8)
Solving equations (6.6) and (6.8) we find the co-ordinates of H2 are H2((f + g – h), (f + g), 0). Similarly the co-ordinates of H3 – H6 are H3(0, g + h, g + h – f), H4(0, g + h – f, g + h), H5(h + f – g, 0, h + f), H6(h + f, 0, h + f – g). 7. The Conic of Orthocentres Using the co-ordinates obtained in Section 6 we find the equation of the conic of orthocentres to be of the form (4.1) with u = 2(f + g)(g + h)(h + f)(f + g – h)(g + h – f)(h + f – g), with v = w = u and (7.1) 2 2 l = – (f + g)(f + g – h)(h + f – g)(h + f)(f – 2f(g + h) + 2(g + h) ), (7.2) where m, n may be obtained from l by cyclic change of f, g, h. 8. The Equations of the Six Euler Lines From the co-ordinates of the circumcentres and orthocentres we may obtain the equations of the Euler lines. The equation of O1H1 is (f + g – h)x – (f + g)y – (3f + 3g – h)z = 0, (8.1) and the equation of O2H2 is (f + g)x – (f + g – h)y + (3f + 3g – h)z = 0. Their point of intersection 12 has co-ordinates 12(3f + 3g – h, 3f + 3g – h, – h).
(8.2) (8.3)
The points of intersection 34, 56 have co-ordinates that may be obtained from those of 12 by cyclic change of x, y, z and f, g, h.
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The equation of O3H3 is (3g + 3h – f)x – (g + h – f)y + (g + h)z = 0. The point of interaction of O2H2 and O3H3 is the point 23, which has co-ordinates 23(3f2 + f(g – 3h) – 2g(g + h), 3f2 – f(5g + 9h) – (g + h)(8g – 3h), – (f(2g + 3h) + (g + h)(2g – 3h))
(8.4)
(8.5)
The points of intersection 45, 61 have co-ordinates that may be obtained from those of 23 by cyclic change of x, y, z and f, g, h. 9. The Euler Conic through 12, 23, 34, 45, 56, 61 From the co-ordinates derived in Section 8 we may obtain the equation of the Euler Conic which is of the form (4.1) with u = 2(f – 3(g + h))((2f2 + f(g + h) – (g – h)2), (9.1) where v and w may be obtained from u by cyclic change of f, g, h, and l = 2f3 + 14f2(g + h) + f(7g2 + 16gh + 7h2) – 5g3 + 9g2h + 9gh2 – 5h3,
(9.2)
and where m, n may be obtained from l by cyclic change of f, g, h. 10. Further Comments Cabri indicates that the six in-centres are also co-conic and also the six nine-point centres, but not other main triangle centres such as the centroids or symmedian points. It may also be observed that the hexagon formed by the contiguous points of intersection of the Euler lines has opposite sides parallel.
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Article: CJB/2011/134
When the Cevian Conic is a Circle Christopher Bradley Abstract: When the Cevian point is Gergonne’s point the Cevian Conic is a Circle with centre at the incentre.
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Fig. 1 The Cevian Circle and the line of centres Ge I J deL 1. Introduction Triangle ABC is given with the Cevian point Ge, Gergonne’s point. The feet of the Cevians are labelled D, E, F. Circles EFGe, FDGe, DEGe are now drawn to produce the other six points of 1
intersection of these circles with the sides of ABC. See Fig. 1 for the labelling of these points. From Article 133 we know that these six points lie on a conic, but Cabri plus II indicates that in this case alone the conic is a circle. This is proved in the following sections using areal coordinates. The centre of this circle is the incentre of ABC. The lines LU, MV, NW create a triangle A'B'C' and it is shown that the centre J of the circumcircle of triangle A'B'C' lies on the line Ge I (which also contains deL according to Kimberling’s list of triangle centres). This we are able to verify. The point J appears to be a new triangle centre. 2. The points D, E, F and the circle Ge E F Ge has co-ordinates Ge(1/(b + c – a), 1/(c + a – b), 1/(a + b – c)) so D, E, F have co-ordinates D(0, 1/(c + a – b), 1/(a + b – c)), E(1/(b + c – a), 0, 1/(a + b – c)), F(1/(b + c – a), 1/c + a – b), 0). In areals the equation of a circle has the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
Here u, v, w depend on the co-ordinates of three points lying on the circle. In the case of circle Ge E F we find u = – (1/k)(2a4 – 5a3(b + c) + a2(3b2 + 10bc + 3c2) + a(b3 – 5b2c – 5bc2 + c3) – (b2 – c2)2)), (2.2) 4 3 2 2 2 2 2 v = – (1/k)(2a – 5a b + a (b – c)(3b + 4c) + ab(b – c) – (b – c) (b + bc – 2c )), (2.3) 3 2 2 2 w = – (1/k)((a + b – c)(2a – a (2b + 3c) + 2ab(c – b) + (b – c)(2b – bc – c )), (2.4) where k = 2(a2 + b2 + c2 – 2ab – 2bc – 2ca).
(2.5)
3. The points U, V, W, L, M, N The circle Ge E F meets AB, z = 0, at the point U with co-ordinates (x, y, z), where x = – 2 (c + a – b)(2a2 – b2 + 2c2 – bc – 4ca – ab), y = 2(b + c – a)(– 2a2 + b2 + c2 – 2bc + ca + ab), z = 0.
(3.1) (3.2) (3.3)
The points of co-ordinates V and W follow from Equations (3.1) – (3.3) by cyclic change of x, y, z and a, b, c. The circle Ge E F meets CA, y = 0, at the point L with co-ordinates (x, y, z), where x = – 2(a + b – c)(2a2 + 2b2 – c2 – bc – ca – 4ab), y = 0, z = 2(b + c – a)(– 2a2 + b2 + c2 – 2bc + ca + ab). 2
(3.4) (3.5) (3.6)
The points of co-ordinates M and N follow from Equations (3.4) – (3.6) by cyclic change of x, y, z and a, b, c. 4. Circle LMNUVW This has an equation of the form (2.1), where we find u = (1/m)(2a6 – 3a5(b + c) – 2a4(3b2 – 4bc + 3c2) + 2a3(7b3 – b2c – bc2 + 7c3) – 2a2(3b4 + 6b3c – 10b2c2 + 6bc3 + 3c4) – a(3b5 – 13b4c + 10b3c2 + 10b2c3 – 13bc4 + 3c5) + (2b6 – 4b5c – 2b4c2 + 8b3c3 – 2b2c4 – 4bc5 + 2c6)), (4.1) in which m = (a2 + b2 + c2 – 2ab – 2bc – 2ca)2.
(4.2)
Values of v and w follow from Equation (4.1) by cyclic change of a, b, c. 5. Lines LU, MV, NW and points A', B', C' The equation of the line LU is (a – b – c)(2a2 – a(b + c) – (b – c)2)x + (a – b + c)(2a2 – a(b + 4c) – b2 – bc + 2c2)y + (a + b – c)(2a2 – a(4b + c) + 2b2 – bc – b2)z = 0. (5.1) The equations of MV and NW follow from Equation (5.1) by cyclic change of x, y, z and a, b, c. The point A' = MV^NW has co-ordinates which can now be evaluated and they are (x, y, z), where x = – 3a(a + b – c)(c + a – b), (5.2) 2 2 y = (a + b – c)(a + a(b + c) – 2(b – c) ), (5.3) 2 2 z = (c + a – b)(a + a(b + c) – 2(b – c) ). (5.4) The co-ordinates of B' and C' may now be written down from Equations (5.2) – (5.4) by cyclic change of x, y, z and a, b, c. 6. Centre of circle UVWLMN The centre of the conic with equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0
(6.1)
has co-ordinates (x, y, z), where x = vw – gv – hw – f2 + fg + hf, y = wu – hw – fu – g2 + gh + fg,
(6.2) (6.3)
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z = uv – fu – gv – h2 + hf + gh.
(6.4)
The equation of circle UVWLMN appears in Section 4 and after some lengthy calculations it is found that (x, y, z) ≈ (a, b, c), the incentre. 7. Circle A'B'C' and its centre J Circle A'B'C' has an equation of the form (6.1), where u = (b + c – a)2(a2 + a(b + c) – 2(b – c)2)(5a2 – a(b + c) – 4(b – c)2),
(7.1)
with v and w following from Equation (7.1) by cyclic change of a, b, c, and 2f = (a + b – c)(c + a – b)(17a4 – 38a3(b + c) + 2a2(6b2 + 31bc + 6c2) + 22a(b3 – b2c – bc2 + c3) – (13b4 + 2b3c – 30b2c2 + 2bc3 + 13c4)),
(7.2)
and 2g, 2h follow from Equation (7.2) by cyclic change of a, b, c. The co-ordinates (x, y, z) of the centre J of this circle may now be calculated using Equations (6.2) to (6.4) and we find x = 7a3 – 5a2(b + c) + a(b – c)2 – 3(b3 – b2c – bc2 + c3) (7.3) and y, z follow from Equation (7.3) by cyclic change of a, b, c. The equation of I Ge may be calculated and has equation (b + c – a)2(b – c)x + (c + a – b)2(c – a)y + (a + b – c)2(a – b)z = 0.
(7.4)
It may now be checked that J lies on the line I Ge, as does deLongchamps point with coordinates (x, y, z), where x = – 3a4 + 2a2(b2 + c2 + (b2 – c2)2, (7.5) and y, z follow from Equation (7.5) by cyclic change of a, b, c. It appears from Kimberling’s list of Triangle Centres that J is a new triangle centre.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/135
Constructing Triangles with Coincident Centroids Christopher Bradley Abstract: Given two triangles with coincident centroids it is shown how to construct an unlimited number of triangles with the same centroid.
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Fig. 1 Triangles with the same centroid 1. Introduction In this short article we are given triangles ABC and LMN with the same centroid G and LMN not the medial triangle of ABC. The lines AL, BM, CN define a third triangle UVW and lines AU, BV, CW define a fourth triangle RST. Lines LR, MS, NT now define a fifth triangle XYZ. It is shown that the centroids of the three additional triangles coincide with the centroids of the first two. This procedure of creating new triangles may be continued indefinitely. In retrospect the results are obvious enough, but seemed interesting on discovery. 1
2. The analysis Using areal co-ordinates we take ABC to be the triangle of reference and L, M, N to have coordinates L(0, p, 1 – p), M(1 – p, 0, p), N((p, 1 – p, 0), it being trivial that LMN also has G(1, 1, 1) as centroid. AL, BM, CN have equations (p – 1)y + pz = 0, px + (p – 1)z = 0 ,
(p – 1)x + py = 0.
(2.1)
U = BM^CN has co-ordinates U(p(1 – p), (1 – p)2, p2). V = CN^AL has co-ordinates V(p2, p(1 – p), (1 – p)2). W = AL^BM has co-ordinates W((1 – p)2, p2, p(1 – p)). The centroid of triangle UVW clearly has G as centroid. AU, BV, CW have equations (1 – p)2z = p2y, (1 – p)2x = p2z,
(1 – p)2y = p2x.
(2.2)
R = BV^CW has co-ordinates (p2(1 – p)2, p4, (1 – p)4). S = CW^AU has co-ordinates ((1 – p)4, p2(1 – p)2, p4). T = AU^BV has co-ordinates (p4, (1 – p)4, p2(1 – p)2). The centroid of triangle RST clearly has G as centroid. LR has equation p(1 – p)(2p3 – 3p2 + 3p – 1)x + p2(1 – p)2(pz – (1 – p)y) = 0.
(2.3)
MS, NT have equations that may be written down from Equation (2.3) by cyclic change of x, y, z. X = MS^NT has co-ordinates (x, y, z), where x = (3p6 – 9p5 + 18p4 – 21p3 + 15p2 – 6p + 1, y = p2(p – 1)(3p3 – 6p2 + 6p – 2), z = p(1 – p)2(3p3 – 3p2 + 3p – 1).
(2.4) (2.5) (2.6)
Y = NT^LR and Z = LR^MS have co-ordinates that may be written down from Equations (2.4) – (2.6) by cyclic change of x, y, z.
2
Again it is evident that triangle XYZ has centroid G and the way the analysis proceeds makes it evident that the centroid of subsequent triangles formed from previous ones have the property that their co-ordinates are obtained from one vertex by cyclic change of x, y, z and therefore all have centroids at G.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/136
Properties of the Symmedian points in a Cyclic Quadrilateral Christopher Bradley Abstract: If ABCD is a cyclic quadrilateral then the symmedian points of triangles ABC and DBC are shown to have some remarkable properties. Many pairs of lines may be constructed whose intersections lie on the sides of the diagonal point triangle. P
Z
To P To Z
G A
U
da D
Q
V
O
M
H S
B
R
I T L
N
bc
C
To E E
jk
K
J Y
X
W
To F
F
Fig. 1 1
A cyclic quadrilateral ABCD and symmedian points S and T 1. Introduction As there are no metrical properties to be proved and no pairs of parallel lines involved we use projective homogeneous co-ordinates in the proofs, as these are technically easier to use. It must be emphasized that the symmedian point of a triangle inscribed in a conic is not the same point as the symmedian point of the same triangle when inscribed in a circle (unless the conic is a circle). We describe how the figure above was constructed, mentioning properties to be proved as we proceed. The quadrilateral ABCD is drawn inscribed in a conic Σ. S and T are constructed as the symmedian points of triangles ABC and DBC respectively. This first requires drawing the tangents at A, B, C, D and, for example, if those at B and C meet at F then AF passes through S and DF passes through T and so on. In the figure the tangent at A passes through U and V, the tangent at B passes through F, U and Z, the tangent at C passes through F, W and V and the tangent at D passes through Z and W. The intersection of the diagonals AC and BD is labelled O, AB and CD meet at G, AD and BC meet at E. Triangle OGE is the diagonal point triangle. We then constructed the following points on the conic Σ: Q, R, M, N, J, K and X, Y. Q = CT^Σ, R = BS^Σ, M = CS^Σ, N = BT^Σ, J = DS^Σ, K = AT^Σ, X = AS^Σ, Y = DT^Σ. The following properties hold and proofs are given: (i) QR, MN, ST, JK, XY, ZGV, UOW all pass through E; (ii) The following points lie on the line joining the diagonal points O and G: P = BQ^CR, jk = BK^CJ, bc = AY^DX, L = BN^CM, I = CQ^BR, H = AK^DJ. The tangents at A and D do, of course, lie on this line also. It would have been possible to construct the symmedian points of triangles ABD and CBD and then the joins of pairs of these points pass through the three diagonal points. However, the figure is overloaded as it is and this final observation is left to the imagination. 2. The tangents at A, B, C, D and the points F, U, V, Z, W, S and T, E and line ST We take the conic Σ to have equation fyz + gzx + hxy = 0,
(2.1)
where, without loss of generality, we may choose h = – f – g, so that the conic passes through 2
A(1, 0, 0), B(0, 1, 0), C(0, 0, 1) and D(1, 1, 1). The ratio of f : g depends on a fifth point on the conic that we do not need to specify. At a certain stage of the proof we stop using h and replace it by – (f + g). The tangents tA, tB, tC, tD to Σ at the points A, B, C, D are respectively as follows: hy + gz = 0, hx + fz = 0, gx + fy = 0 and (g + h)x + (h + f)y + (f + g)z = 0. Tangents tB and tC meet at F(– f, g, h). Tangents tC and tA meet at V(f, – g, h). Tangents tA and tB meet at U(f, g, – h). The equations of BV and CU are respectively hx – fz = 0 and fy – gx = 0 and these meet at the symmedian point S of triangle ABC with co-ordinates S(f, g, h). Tangents tA and tD meet at da (f(g – h), – g(g + h), h(g + h)). Tangents tB and tD meet at Z (– f(f + h), g(f – h), h(f + h)). Tangents tC and tD meet at W(f(f + g), – g(f + g), h(g – f)). The equations of CZ and BW are respectively g(h – f)x – f(f + h)y = 0 and h(g – f)x – f(f + g)z = 0 and these meet at the symmedian point T of triangle DBC with co-ordinates T(– f(f + g)(f + h), g(f + g)(f – h), h(f – g)(f + h)). Sides BC, AD meet at E(0, 1, 1). Line ST has equation gh(g – h)x + hf(f + h)y – fg(f + g)z = 0.
(2.2)
and using f + g + h = 0 it may be checked that ST passes through E. 3. Points X, Y, M, N, R, S, J, K on the circumconic Line AF has equation gz = hy and it meets the conic Σ at the point X(– f, 2g, 2h). Line DF has equation (h – g)x – (h + f)y + (f + g)z = 0 and it meets the conic Σ at Y(f, – f – 2g, f + 2g). Line XY has equation 2(f + 2g)(g + h)x + f(f + 2g + 2h)y + f2z = 0
(3.1)
and using, f + g + h = 0, it may be checked that XY passes through E (0, 1, 1). At this point we eliminate h, using h = – f + g and we now use S(f, g, – f – g), T(f, 2f + g, f – g) with ST having equation (f + 2g)x – f(y – z) = 0. Line CS has equation fy = gx and it meets the conic Σ at the point M(2f, 2g, f + g). Line BT has equation (f – g)x = fz and it meets the conic Σ at the point N(2f, f – g, 2(f – g)). 3
Line MN has equation (g – f)x = 2f(y – z), which clearly passes through E(0, 1, 1). Line CT has equation fy = (2f + g)x and it meets the conic Σ at the point Q(2f, 4f + 2g, 2f + g). Line BS has equation (f + g)x + fz = 0 and it meets the conic Σ at the point R(– 2f, g, 2(f + g)). Line QR has equation (2f + g)x = 2f(y – z), which clearly passes through the point E(0, 1, 1). Line DS has equation (2f + g)y + (g – f)z = (f + 2g)x and it meets the conic Σ at the point J with co-ordinates J(f(g – f)(2f + g), g(f – g)(f + 2g),(f + g)(f + 2g)(2f + g)). Line AT has equation (g – f)y + (2f + g)z = 0 and it meets the conic Σ at the point K with coordinates (f(f – g)(2f + g), 2(f2 + fg + g2)(2f + g), 2(f2 + fg + g2)(f – g)). Line JK has equation 2(f + 2g)(f2 + fg + g2)x + f(g – f)(2f + g)(y – z) = 0,
(3.2)
again which clearly passes through the point E(0, 1, 1). 4. The point G and the line GE contains V and Z The point G is the intersection of AB and CD and therefore has co-ordinates G(1, 1, 0). The equation of the line GE is x – y + z = 0. It may now be checked that the points V and Z lie on GE. Similarly it may be shown that FB passes through U and Z. 5. The point P, the line PGF and many points lying on this line The line JQ has equation 2(f2 + fg + g2)x – 2f(f – g)y + f(2f + g)z = 0.
(5.1)
The line KR has equation (f + 2g)(2f + g)x – f(2f + g)y – 2f(g – f)z = 0.
(5.2)
The point P is the intersection of these two lines and therefore has co-ordinates P(f(g – 4f), – 4f2 – 3fg – 2g2, 2g(2f + g)). It may now be shown that P lies on the line FG with equation x = y + z. Diagonals BD and AC meet at the diagonal point O with co-ordinates (1, 0, 1), which clearly lies on FG. 4
The line AY has equation y + z = 0 and the line DX has equation y + z = 2x. These lines meet at the point bc (0, 1, – 1), which lies on both BC and FG. The line CJ has equation g(f + 2g)x + f(2f + g)y = 0. (5.3) The line BK has equation 2(f2 + fg + g2)x = f(2f + g)z.
(5.4)
These two lines intersect at the point jk (f(2f + g), – g(f + 2g), 2(f2 + fg + g2)) and it may now be checked that jk lies on the line FG. The line CM has equation fy = gx and the line BN has equation (f – g)x = fz. These lines meet at the point L (f, g, f – g), which clearly lies on the line FG. The line BR has equation (f + g)x + fz = 0 and the line CQ has equation fy = (2f + g)x and these two lines met at the point I (f, 2f + g, – f – g), which again clearly lies on FG. The line AK has equation (f – g)y = (2f + g)z and the line DJ has equation (f + 2g)x – (2f + g)y + (f – g)z = 0.
(5.5)
These two lines meet at the point H(3f, 2f + g, f – g), which again clearly lies on FG. Finally it may be checked that da, the intersection of the tangents at D and A also lies on FG.
Flat 4, Terrill Court, 12 - 14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2011/137
Affine Transform of the properties of the Euclidean Triangle Christopher Bradley
C'
A
F
B'
N' N
W'
F' O V
U
E'
M V' M' K E H G ' W U'
B
L
D
L' D'
A'
Fig. 1 Affine transform of the Triangle and its Conics
1
C
1. Introduction Given a triangle ABC and a point K a configuration is constructed containing a triplicate ratio conic, a circumconic, a nine-point conic and a 7 pt. conic. When K is actually the symmedian point K(a2, b2, c2) then these conics are the familiar circles associated with a triangle in the Euclidean plane. In this article we take K, a point not on the sides of ABC, to be an arbitrary point K(f, g, h) and what emerges is an affine map of the Euclidean plane. An analytic account of the process is given and an additional conic is discovered (new to me, but probably not original). It should be added that an affine transformation preserves parallels and ratios of segments on a straight line (including midpoints) but not perpendiculars. 2. The Triplicate Ratio Conic A point K is chosen to have areal co-ordinates (f, g, h). Following the construction of the triplicate ratio circle lines are drawn parallel to the sides, producing six points, two on each side. Starting from B and working around the triangle in an anticlockwise direction these points are labelled D, D', E, E', F, F'. Their co-ordinates are easily obtained and are D(0, f + g, h), E(f, 0, g + h), F(h + f, g, 0), D'(0, g, h + f), E'(f + g, 0, h), F'(f, g + h, 0). Any conic in areal co-ordinates has an equation of the form ux2 + vy2 + wz2 + 2pyz + 2qzx + 2rxy = 0,
(2.1)
where u, v, w, p, q, r are constants to be determined from the co-ordinates of five points lying on the conic. Using the above points it is found that they all lie on a conic, which (for obvious reasons) we designate the triplicate ratio conic. The constants for this conic are: u = 2gh(g + h), v = 2hf(h + f), w = 2fg(f + g), (2.2) and p = – f(f2 + f(g + h) + 2gh), q = – g(g2 + g(h + f) + 2hf), r = – h(h2 + h(f + g) + 2fg).
(2.3)
The centre (x, y, z) of a conic in the form of Equation (2.1) is given by x = vw – qv – rw – p2 + pq + pr,
(2.4)
with y, z to be obtained from Equation (2.4) by cyclic change of u, v, w and p, q, r. The centre X of the triplicate ratio conic therefore has co-ordinates (x, y, z), where x = f(f(g + h) + 2gh – f2), y = g(g(h + f) + 2hf – g2) (2.5) 2 z = h(h(f + g) + 2fg – h ). 2
3. The circumcentre O and the circumconic The line KX is now the affine map of the Brocard axis so that the co-ordinates of the circumcentre O are given by O = 2X – K as in the Euclidean plane. After some calculation and cancellation we find O has co-ordinates O(f(g + h – f), g(h + f – g), h(f + g – h)). For the circumconic we know that since it passes through A, B, C it must have an equation of the form pyz + qzx + rxy = 0. (3.1) What we also know is that it passes through points P, Q, R where P = 2O – A, Q = 2O – B, R = 2O – C (the images of A, B, C when rotated by 180o about O). Again after some calculation and even more cancellation we find p = f, q = g, r = z so that the circumconic has equation fyz + gzx + hxy = 0. (3.2) The co-ordinates of O and the equation of the circumconic are as to be expected because we now observe that algebraically what is happening is that a2, b2, c2 are being replaced everywhere by f, g, h respectively. 4. The tangents at A, B, C, the points A', B', C', H and the nine-point conic The equations of the tangents at A, B, C to the circumconic are respectively gz + hy = 0, hx fz = 0, fy + gx = 0. The tangents at B and C meet at the point A'(– f, g, h). Similarly B'(f, – g, h) and C'(f, g, – h). It may now be shown that AA', BB', CC' are concurrent at K. This is the main property of a symmedian point and justifies what has pre-ceded in Section 3. The centroid of a triangle is G with co-ordinates G(1, 1, 1) (whether in the Euclidean or an Affine plane) and using GH = 2OG we may find the co-ordinates of H (which acts as orthocentre) and as expected H has co-ordinates H(1/(g + h – f), 1/(h + f – g), 1/(f + g – h)). The midpoints of the sides L, M, N have co-ordinates L(0, 1, 1), M(1, 0, 1), N(1, 1, 0). The equation of AH is (f + g – h)z = (h + f – g)y
(4.1)
This line meets BC at the point L' with co-ordinates (0, f + g – h, h + f – g). Co-ordinates of M' and N' may be written down by cyclic change of f, g, h and x, y, z.
3
The nine-point conic passes through L, M, N and L', M', N' and its equation is of the form (2.1) with u = g + h – f, v = h + f – g, w = f + g – h, p = – f, q = – g, r = – h. (4.2) The Brocard point Ω is the point of concurrence of the line through A parallel to FD, the line through B parallel to DE and the line through C parallel to EF. It may now be checked that Ω has co-ordinates (1/g, 1/h, 1/f). Similarly Ω' has co-ordinates (1/h, 1/f, 1/g). The 7 pt. Conic has equation ghx2 + hfy2 + fgz2 – f2yz – g2zx – h2xy = 0.
(4.3)
5. The nine points of intersection of triangles DEF and D'E'F' and a conic and a line The point of intersection FD^E'F' has co-ordinates (– f(f + g)(h + f), gh(f + g), gh(h + f)) and the points of intersection DE^F'D' and EF^D'E' lie on the line with equation gh(g + h)x + hf(h + f)y + fg(f + g)z = 0. (5.1) It may be shown that this is the polar of K with respect to the triplicate ratio conic. The point W' = FD^E'F' has co-ordinates ((f + g)h + f), (f + g)(g + h), h2). The co-ordinates of points U', V' may be written down by cyclic change of x, y, z and f, g, h. The point V = FD^F'D' has co-ordinates (f2(h + f), f2(g + h) + hf(g + h) + gh2, h2(h + f)). The coordinates of W and U may be written down by cyclic change of x, y, z and f, g, h. It may now be shown that these six points lie on a conic, which we call the inner conic. Note that in the Euclidean plane this conic is not a circle. Its equation has the form of (2.1) with u = 2(g + h)(f(g + h) + g2 + h2), v = 2(h + f)(g(h + f) + h2 + f2), w = 2(f + g)(h(f + g) + f2 + g2), p = f3 – f2(g + h) – 2f(g2 + gh + h2) – 2gh(g + h) and where q and r may be obtained from p by cyclic change of f, g, h. The centre of the inner conic has co-ordinates x = 3f2 + 2(fg + gh + hf), with y and z obtained by cyclic change of f, g, h.
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5
Article: CJB/2011/138
Explaining some Collinearities amongst Triangle Centres Christopher Bradley Abstract: Collinearities are preserved by an Affine Transformation, so that for every collinearity of triangle centres, there is another when an affine transformation has taken place and vice-versa. We illustrate this by considering what happens to three collinearities in the Euclidean plane when an affine transformation takes the symmedian point into the incentre.
C'
9 pt circle C''
A
Anticompleme ntary triangle
B''
F B'
W' W
OL
F'
X Z
K'' Y G T N' D
B
U
M K 4 H
D'U'
E' V E V'
C
Triplicate ratio Circle
A'' A'
Fig.1 Three collinearities amongst triangle centres in the Euclidean plane 1
1. When the circumconic, the nine-point conic, the triplicate ratio conic and the 7 pt. conic are all circles The condition for this is that the circumconic is a circle and then the tangents at A, B, C form a triangle A'B'C' consisting of the ex-symmedian points A'(– a2, b2, c2) and similarly for B', C' by appropriate change of sign and when this happen AA', BB', CC' are concurrent at the symmedian point K(a2, b2, c2). There are two very well-known collinearities and one less well-known, and we now describe them. First there is the Euler line containing amongst others the circumcentre O with x-co-ordinate a2(b2 + c2 – a2), the centroid G(1, 1, 1), the orthocentre H with x-co-ordinate 1/(b2 + c2 – a2) and T, the nine-point centre which is the midpoint of OH. The equation of the Euler line is (b2 – c2)(b2 + c2 – a2)x + (c2 – a2)(c2 + a2 – b2)y + (a2 – b2)(a2 + b2 – c2)z = 0. (1.1) Secondly there is the Brocard axis which serves as diameter for the 7 pt. circle, with O and K at opposite ends of the diameter. Other points on this axis are X, the centre of both the triplicate ratio circle and the 7 pt. circle, Z the midpoint of the segment joining the Brocard points Ω (1/b2, 1/c2, 1/a2) and Ω'(1/c2, 1/a2, 1/b2), as well as the fourth power point 4 with co-ordinates (a4, b4, c4). Z has co-ordinates (a2(b2 + c2), b2(c2 + a2), c2(a2 + b2)). The equation of the Brocard axis is b2c2(b2 – c2)x + c2a2(c2 – a2)y + a2b2(a2 – b2)z = 0. (1.2) Thirdly there is the line joining K and G. This passes through K''(b2 + c2 – a2, c2 + a2 – b2, a2 + b2 – c2), which is the symmedian point of the anticomplementary triangle (which has ABC as its medial triangle) and the midpoint Y of KK'' (Y is X38 in the Encyclopaedia of Triangle Centres). Its co-ordinates are (b2 + c2, c2 + a2. a2 + b2). The equation of this axis is (b2 – c2)x + (c2 – a2)y + (a2 – b2)z = 0. (1.3) 2. What happens when K is replaced by the incentre I This is brought about by choosing a circumconic (i.e. one that passes through A, B, C) such that the tangents at the vertices form a triangle with vertices at the excentres I1, I2, I3 and then the lines AI1, BI2, CI3 are concurrent at the incentre I. See Fig.2 below for the diagram. From Article 137 what this amounts to is to replace a2, b2, c2 by a, b, c in areal co-ordinates of points and in equations of lines. We are therefore in a position to provide easily the comparison to the situations described in Section 1.
2
I3
7 pt conic A F Feu
W' W F' 9 pt conic
Mi XZ K Na Sp G I T ' Ge
E' V
I2
E V'
Inconic D U
B
D' U'
C
I1
Fig. 2 The image of the Euclidean plane when K is replaced by I First consider the image of the Euler line. This contains the points Mi (the Mittenpunkt) with coordinates (a(b + c – a), b(c + a – b), c(a + b – c)), replacing O, the centroid G(1, 1, 1) and Ge (Gergonne’s point) with co-ordinates (1/(b + c – a), 1/(c + a – b), 1/(a + b – c)), replacing H and the midpoint of OH is now replaced by the midpoint of MiGe (still labelled T in Fig. 2). This, of course, is a known collinearity. The equation of this axis is (b – c)(b + c – a)x + (c – a)(c + a – b)y + (a – b)(a + b – c)z = 0.
3
(2.1)
Next consider the image of the Brocard axis. This has a diameter MiI and the 7 pt. conic has centre the image of X. The images of the Brocard points denoted by ω and ω' in Fig. 2 have coordinates (1/b, 1/c, 1/a) and (1/c, 1/a, 1/b) respectively. The midpoint of ωω' (still labelled Z) has co-ordinates (a(b + c), b(c + a), c(a + b)) and is X37 in the Encyclopaedia. It also contains K, the symmedian point, which is the image of 4, the fourth power point. The equation of this axis is bc(b – c)x + ca(c – a)y + ab(a – b)z = 0. (2.2) Finally let us consider the image of the third line in Section 1. This is the line IG, which also contains Sp(the image of Y) with co-ordinates (b + c, c + a, a + b) and Na(the image of K'') with co-ordinates (b + c – a, c + a – b, a + b – c). Note that Sp is the midpoint of INa. The equation of this axis is (b – c)x + (c – a)y + (a – b)z = 0. (2.3) Admittedly there is nothing new in this short article, but it serves to illustrate that there is a system and reason for many of the collinearities appearing in the Encyclopaedia of Triangle Centres.
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4
Article: CJB/2011/139
The Harmonic Cevian Conic Christopher Bradley Abstract: If D, E, F are the feet of Cevians with D on BC etc. and if L is the harmonic conjugate of C with respect to B and D and U is the harmonic conjugate of B with respect to C and D and if M, N, V, W are similarly defined then a conic passes through the six points L, U, M, V, N, W. There is one internal Cevian point for which the conic is a circle.
A
N V F W
E P
M
D B
L
U
C
Fig. 1 The Harmonic Cevian Conic 1. Deriving the equation of the Harmonic Cevian Conic In areal co-ordinates the equation of a conic is of the form px2 + qy2 + rz2 + 2uyz + 2vzx + 2wxy = 0.
(1.1)
Here p, q, r, u, v, w are constants to be determined by the co-ordinates of five points lying on the conic. Suppose P has co-ordinates (f, g, h), then D, E, F have co-ordinates D(0, g, h), E(f, 0, h), F(f, g, 0). U is the harmonic conjugate of B with respect to C and D and therefore has co-ordinates U(0, g, 2h) and L has co-ordinates L(0, 2g, h). Similarly V, W, M, N have co-ordinates V(2f, 0, h), W(f, 2g, 0), M(f, 0, 2h), N(2f, g, 0). Inserting these co-ordinates into Equation (1.1) and solving we find p = 4g2h2, q = 4h2f2, r = 4f2g2, u = – 5f2gh, v = – 5fg2h, w = – 5fgh2.
(1.2)
This conic is a circle if and only if f2(4g2 + 10gh + 4h2) = a2, g2(4h2 + 10hf + 4f2) = b2, h2(4f2 + 10fg + g2) = c2.
(1.3)
Cabri II plus indicates there are indeed four points P which gives rise to a circle, one internal point and three external points. See Fig. 2.
A
D B
U
L
C
W P
M
F
E
N
Fig, 2 One of the Harmonic Cevian Circles
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Article: CJB /2011/140
The Centroid of Centroids Christopher Bradley Abstract: Let P be a point not on the sides of a triangle and suppose the centroids of triangle PBC, PCA, PAB are L, M, N respectively. Now let J be the centroid of triangle LMN. Then J lies on line PG, where G is the centroid of triangle ABC. The point of concurrence Q of AL, BM, CN also lies on PG.
A
M
N G Q J P
L C
B Fig. 1 The line of Centroids
1.4458790849 1. The points L, M, N and the centroid J of trianglecm LMN Result: 1.5000000000 Let P have areal co-ordinates (f, g, h). Then L, the centroidcm of triangle PB, C has co-ordinates (x, 0.9639193899 y, z), where x = (1/3)(f/(f + g + h)), (1.1) y = (1/3)(g/(f + g + h)) + 1/3, (1.2)
z = (1/3)(h/(f + g + h)) + 1/3,
(1.3)
Points M and N have co-ordinates that may be written down from Equations (1.1) to (1.3) by cyclic change of x, y, z and f, g, h. It follows that the centroid J of triangle LMN has co-ordinates (x, y, z), where x = (5f + 2(g + h))/(9(f + g + h)), y = (5g + 2(h + f))/(9(f + g + h)), z = (5h + 2(f + g))/(9(f + g + h)).
(1.4) (1.5) (1.6)
2. The line PG contains J The co-ordinates of G, the centroid of triangle ABC are G(1, 1, 1) so the equation of PG is (g – h)x + (h – f)y + (f – g)z = 0. (2.1) It may now be checked that the point J with co-ordinates given by Equations (1.4) – (1.6) lies on PG. 3. The lines AL, BM, CN and their point of concurrence Q The equation of the line AL is (f + g + 2h)y = (f + 2g + h)z.
(3.1)
The equations of BM and CN follow by cyclic change of x, y, z and f, g, h. These three lines met at the point Q with co-ordinates Q(2f + g + h, f + 2g + h, f + g + 2h). It may now be checked that Q lies on the line PG. And so P, J, Q, G are collinear. It may also be checked that PJ = (2/3)PG.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
Article: CJB/2011/141
Three Centroids created by a Cyclic Quadrilateral Christopher Bradley Abstract: The centroid of the quadrilateral considered to be an area of constant density is G, the centroid of the quadrilateral considered as having unit masses at its vertices is N, the centroid of the quadrilateral considered as having unit masses at its vertices and mass of two units at E (the intersection of its diagonals) is F. It is shown that E, F, N, G are collinear.
A
J
G12 Gd
B
M
Gc G41
G N
Gb F
K
G23
E Ga
G34
D
L C Fig. 1 Centroids in a Cyclic Quadrilateral 1. Introduction In the diagram above ABCD is a cyclic quadrilateral in which the diagonals meet at the point E. Points Ga, Gb, Gc, Gd are the centroids of triangles BCD, ACD, ABD, ABC respectively. GaGc meets GbGd at the point G. Points J, K, L, M are the midpoints of AB, BC, CD, DA respectively. 1
JL meets KM at the point N. Points G12, G23, G34, G41 are the centroids of triangles ABE, BCE, CDE, DAE respectively. G12G34 meets G23G41 at F. The centroid of the quadrilateral considered to be an area of constant density is G, the centroid of the quadrilateral considered as having unit masses at its vertices is N, the centroid of the quadrilateral considered as having unit masses at its vertices and a mass of two units at E (the intersection of its diagonals) is F. It is shown that E, F, N, G are collinear. We use oblique axes with x-axis in the direction EA and with y-axis in direction EB. Thus E has co-ordinates (0. 0) and A, B, C, D may be chosen to have co-ordinates (fg, 0), (0, hf), (– hk, 0), (0, – gk), where f, g, h, k are any constants and the co-ordinates reflect the relation AE x EC = BE x ED implying, by the intersecting chord theorem, that ABCD is cyclic. 2. The centroid G of ABCD, considered to be an area of uniform density The centroid Gd of triangle ABC has co-ordinates (1/3)(fg – hk, fh). The centroid Gb of triangle ACD has co-ordinates (1/3)(fg – hk, – gk). The centroid Gc of triangle ABD has co-ordinates (1/3)(fg, fh – gk). The centroid Ga of triangle BCD has co-ordinates (1/3)(– hk, fh – gk). Since ABC and ACD combine to give ABCD it follows that the centroid of ABCD lies on GdGb. Similarly it lies on GcGa. It follows immediately, by inspection, that G = GdGb^GcGa has coordinates (1/3)(fg – hk, fh – gk). The line EG has equation (fh – gk)x = (fg – hk)y. (2.1) 3. The centroid F when there is a mass 1 at each of A, B, C, D and a mass 2 at E These masses are arranged so that we may suppose there are equal masses at the vertices of each of the triangles ABE, BCE, CDE, DAE. This means that if the centroids of these triangles are G12, G23, G34, G41 respectively then F will be the intersection of the lines G23G41 and G12G34. The centroid G12 of ABE has co-ordinates (1/3)(fg, fh). The centroid G23 of BCE has co-ordinates (1/3)(– hk, fh). The centroid G34 of CDE has co-ordinates (1/3)(– hk, – gk). The centroid G41 of DAE has co-ordinates (1/3)(fg, – gk). The line G23G41 has equation 3(fh + gk)x + 3(fg + hk)y = gh(f2 – k2).
(3.1)
The line G12G34 has equation 3(fh + gk)x – 3(fg + hk)y = fk(g2 – h2).
(3.2)
2
The co-ordinates of F = G23G41^G12G34 are (1/6)(fg – hk, fh – gk). It is immediately obvious that F lies on EG and EG = 2EF. 4. The centroid N when there are equal masses at each of A, B, C, D N has co-ordinates a quarter of the sum of those of A, B, C, D and are therefore (1/4)(fg – hk, fh – gk). If J, K, L, M are the midpoints of AB, BC, CD, DA respectively then N is also the intersection of JL and KM. Evidently N lies on EG and is such that EN = 1.5 x EF.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article: CJB/2011/142
Incircle Conjugation Christopher Bradley Abstract: A construction is presented in which points (f, g, h) in barycentric co-ordinates are transformed into points (a/f, b/g, c/h). This is clearly a conjugation and we call it incircle conjugation. A'
A N F
W
V
E'
M F'
Mi
I
X
B
D'
D
B'
U
Ge
L
Fig. 1 A construction for incircle conjugation 1. Introduction 1
E
C
C'
Given triangle ABC the ellipse through A, B, C with equation ayz + bzx + cxy = 0
(1.1)
is drawn. The method involved is described in Article 137, but with the general point K(f, g, h) replaced by I(a, b, c), the incentre. The centre of this ellipse is the Mittenpunkt Mi with coordinates (a(b + c – a), b(c + a – b), c(a + b – c)). The current article describes a construction, using this ellipse, whereby from a point P (not on the sides of ABC) with co-ordinates (f, g, h), the incircle conjugate Q with co-ordinates (a/f, b/g, c/h) is determined. In Fig. 1 above the incircle conjugate of Mi is Gergonne’s point Ge with co-ordinates (1/(b + c – a), 1/(c + a – b), 1/(a + b – c)). The procedure is as follows: Draw AP to meet the ellipse (1.1) again at R. Through R draw a line parallel to BC to meet the ellipse (1.1) again at R'. Then it transpires that AR' passes through the incircle conjugate Q. Repeating this procedure starting from B and C and we have the result that BS' and CT' all pass through Q and thereby the incircle conjugate Q is obtained. Proof of this result is given in the following Sections. In Fig. 1 the construction shows how Ge is the incircle conjugate of Mi, with R, S, T labelled U, V, W and R', S', T' labelled L, M, N. 2. Analysis of the procedure for obtaining the incircle conjugate Q of a point P Let P have barycentric co-ordinates (f, g, h). Then the equation of AP is hy = gz. AP meets the ellipse with Equation (1.1) again at the point R with co-ordinates (x, y, z), where x = – agh, y = g(bh + cg), z = h(bh + cg). (2.1) The equation of the line through R parallel to BC is (g + h)(bh + cg)x + agh(y + z) = 0.
(2.2)
This line meets the ellipse with Equation (1.1) again at the point R' with co-ordinates (x, y, z), where x = – agh, y = bh(g + h), z = cg(g + h). (2.3) It is immediate that AR' passes through Q with co-ordinates (a/f, b/g, c/h). When P is Mi, then Q is Ge, which acts as a pseudo- orthocentre in the sense that if AGe meets BC at D and the circumconic (1.1) at L then GeD = DL and similarly for BGe and CGe. Proof of this is left to the reader. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
2
Article: CJB/2011/143
The Nine Point Circle of the Diagonal Point Triangle Christopher Bradley Abstract: An analysis is conducted of the ten point rectangular hyperbola through the mid-points of a cyclic quadrilateral and of the nine-point circle of the diagonal point triangle.
S
D
A
G T O
2
5
R
E
X 3
6
C
U
P
8
4 Q
B
1
7
9
F
Fig. 1 A Cyclic Quadrilateral, its Midpoint Conic and associated Circles
1
1. Introduction In the Figure above ABCD is a cyclic quadrilateral, centre O. AC meets BD at E. AB meets CD at F. AD meets BC at G. So triangle EFG is the diagonal point triangle. Points P, Q, R, S, T, U are respectively the midpoints of AB, BC, CD, DA, AC, BD. An analysis is made of the 10-point rectangular hyperbola that passes through these six midpoints, E, F, G and O. The centre of this rectangular hyperbola is labelled X. Next it is shown that the following sets of points are concyclic: (i) R, F, P, O; (ii) S, G, Q, O; (iii) T, E, U, O. The centres of these circles Σ1, Σ2, Σ3 are labelled 1, 2, 3 respectively. Points 4, 5, 6 are the intersections of circles Σ2, Σ3 and Σ3, Σ1 and Σ1, Σ2 respectively. Points 7, 8, 9 are TU^FG, PR^GE, SQ^EF respectively and TU, PR, SQ all pass through X. Finally points 1, 2, 3, 4, 5, 6, 7, 8, 9, X are concyclic and proves to be the diagonal point triangle’s nine-point circle (David Monk – private communication). In the sections that follow we prove these assertions using areal co-ordinates with ABC as triangle of reference. 2. Points A, B, C, D, E, F, G Points A, B, C have co-ordinates (1, 0, 0), (0, 1, 0), (0, 0, 1) respectively and we take D to have co-ordinates D (– a2t(1 – t), b2(1 – t), c2t). E = AC^BD has co-ordinates E (– a2(1 – t), 0, c2), F = AB^CD has co-ordinates F(– a2t, b2, 0) and G = AD^BC has co-ordinates G (0, b2(1 – t), c2t). 3. Points P, Q, R, S, T, U, O P is the midpoint of AB and has co-ordinates P (1, 1, 0). Q is the midpoint of BC and has coordinates Q (0, 1, 1). R is the midpoint of CD and has co-ordinates R (– a2t(1 – t), b2(1 – t), – a2t(1 – t) + b2(1 – t) + 2c2t). S is the midpoint of DA and has co-ordinates S (– 2a2t(1 – t) + b2(1 – t) + c2t, b2(1 – t), c2t), T is the midpoint of AC and has co-ordinates (1, 0, 1) and U is the midpoint of BD and has co-ordinates U (– a2t(1 – t), – a2t(1 – t) + 2b2(1 – t) + c2t, c2t). The circumcentre O has co-ordinates O(a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)). 4. The Rectangular Hyperbola and its centre X The equation of a conic in areals has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(4.1)
It may now be shown that a rectangular hyperbola passes through O, E, F, G, P, Q, R, S, T, U. It is a rectangular hyperbola because the orthocentre of triangle EFG is the point O. Its equation has the following values for u, v, w, f, g, h. u = – 2b2c2, v = 2c2a2t, w = 2a2b2(1 – t), f = – a2(b2(1 – t) + c2t), g = – a2b2(1 – t) + b2c2, h = c2(b2 – a2t). (4.2) 2
The co-ordinates of the centre X of this rectangular hyperbola are (x, y, z), where x = vw – gv – hw –f2 + fg + hf = – 2a2t(1 – t) + b2(1 – t) + c2t, y = wu – hw – fu – g2 + gh + fg = – a2t(1 – t) + 2b2(1 – t) + c2t, z = uv – fu – gv – h2 + hf + gh = – a2t(1 – t) + b2(1 – t) + 2c2t.
(4.3) (4.4) (4.5)
5. Circles RFPO, SGQO, TEUO A circle in areal co-ordinates has the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(5.1)
It may now be shown that circle Σ1 = RFPO has this form with u = b2c2/(2(a2t – b2)), v = – c2a2/(2(a2t – b2)), w = a2b2(1 – t)/(2(a2t – b2)),
(5.2)
And it may be shown that circle Σ2 = SGQO has the same form but with u = – b2c2/(2(b2( 1 – t) + c2t)), v = – c2a2t/(2(b2( 1 – t) + c2t)), w = – a2b2(1 – t)/(2(b2( 1 – t) + c2t)).
(5.3)
And it may be shown that circle Σ3 = TEUO has the same form but with u = b2c2/(2(a2(1 – t) – c2)), v = c2a2t/(2(a2(1 – t) – c2)), w = – a2b2(1 – t)/(2(a2(1 – t) – c2)).
(5.4)
6. Points 4, 5, 6 Circles Σ2 and Σ3 meet at points O and 4, where 4 has co-ordinates (x, y, z), where x = a2t(c2 – a2(1 – t)), y = b2(c2t + b2(1 – t)), z = c2t(a2 + b2 – c2).
(6.1)
Circles Σ3 and Σ1 meet at points O and 5, where 5 has co-ordinates (x, y, z), where x = – a2t(1 – t)(b2 + c2 – a2), y = b2(1 – t)(b2 – a2t), z = c2t(c2 – a2(1 – t)).
(6.2)
Circles Σ1 and Σ2 meet at points O and 6, where 6 has co-ordinates (x, y, z), where x = a2( 1 – t)(a2t – b), y = – (1 – t)b2(c2 + a2 – b2), z = – c2(c2t + b2( 1 – t)).
(6.3)
7. The Nine-Point Circle of EFG passes through 1, 2, 3, 4, 5, 6, 7, 8, 9,and X The circle Σ passing through points 4, 5, 6 has the form of Equation (5.1) where u = (1/k)(b2c2(a4t(t – 1) + 2a2(c2t – b2(t – 1)) + b4(t – 1) – 2b2c2 – c4t)), (7.1) 2 2 4 2 2 2 4 2 2 4 v = (1/k)(a c t(a t(t – 1) + 2a t(c t – b (t – 1)) + b (t – 1) – 2b c t + c t)), (7.2) 2 2 4 2 2 2 4 2 2 4 w = (1/k)(a b (1 – t)(a t(t – 1) + 2a (t – 1)(b (t – 1) – c t) + b (1 – t) + 2b c (t – 1) – c t)), (7.3)
3
and where k = 4(a2t – b2)(a2(t – 1) + c2)(b2(t – 1) – c2t).
(7.4)
It may now be verified that X lies on this circle. I am indebted to David Monk (private communication) for the following: It is standard that the circumcentre O of a cyclic quadrilateral is the orthocentre of its diagonal triangle EFG. Since RFPO is the circle on OF as diameter its centre 1 is the midpoint of OF; likewise 2, 3. The foot of the altitude OF is on both the circles SQGO and TEUO. This is 4; likewise 5, 6. TU is the join of the midpoints of AC and BD. By the theorem on the midpoints of the diagonals of a complete quadrilateral it passes through the midpoint of FG. This is 7 and likewise 8, 9. Finally X, the centre of a rectangular hyperbola circumscribing EFG lies on the nine-point circle. It follows that the circle through points 4, 5, 6 with the above equation is the nine-point circle of the diagonal point triangle.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/144
Four Nine-Point Circles Christopher Bradley
F
L N A
D
S Y M G
P
U
E X
T
B
R
Z Q C
Fig.1 Nine-Point Circles intersect at a point on the circumcircle of the Diagonal Point Triangle Abstract: In any quadrilateral ABCD the four nine-point circles of triangles BCD, ACD, ABD, ABC have a point in common that lies on the hyperbola through the mid-points and on the circumcircle of the diagonal point triangle. 1
1. Introduction Given a quadrilateral ABCD (with neither pair of opposite sides parallel) in which P, Q, R, S, T, U are the midpoints of sides AB, BC, CD, DA, AC , BD and diagonal point triangle EFG, then a hyperbola Σ passes through the nine points P, Q, R, S, T, U, E, F, G. We also prove that the ninepoint circles of triangles BCD, ACD, ABD, ABC have a common point which lies on Σ and also on the circumcircle of triangle EFG. Areal co-ordinates are used with ABC as triangle of reference. Note that the quadrilateral does not have to be convex and when, for example, D lies in triangle ABC then the diagonal point triangle is the triangle formed by feet of the Cevians of D in triangle ABC. 2. Co-ordinates of points With ABC as triangle of reference we have A(1, 0, 0, B(0, 1, 0), C(0, 0, 1). Let D have coordinates D(f, g, h) with f + g + h = 1. Unnormalised co-ordinates for other points are soon determined and are: P(1, 1, 0), Q(0, 1, 1), R(f, g, 1 + h), S(1 + f, g, h), T(1, 0, 1), U(f, 1 + g, h), E(f, 0, h), F(f, g, 0), G(0, g, h). 3. Conic Σ through the midpoints and E, F, G and its centre X The conic Σ through P, Q, R, S, T, U, E, F, G is of the form ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, where it is easily checked that u = 2gh, v = 2hf, w = 2fg, l = – f(g + h), m = – g(h + f), n = – h(f + g)).
(3.1)
(3.2)
Using the formula x = vw – mv – nw – l2 + lm + nl for the centre X we obtain x = 2f + g + h. Similarly y = f + 2g + h and z = f + g + 2h. 4. The circumcircle of triangle EFG This has equation gh(a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(f + h)))x2 + ... + ... – (g + h)(a2gh(f + g)(h + f) + f(h – g)(b2h(f + g) – c2g(h + f)))yz – ... – ... = 0.
(4.1)
where the dots imply additional terms to be obtained by cyclic change of x, y, z and f, g, h and a, b, c. 5. The nine-point circle of triangle ABC
2
This has equation (b2 + c2 – a2)x2 + (c2 +a2 – b2)y2 + (a2 + b2 – c2z2 – 2a2yz – 2b2zx – 2c2xy = 0.
(5.1)
This meets Σ at the point Y with co-ordinates (x, y, z), where x = f(a2(g – h) – (b2 – c2)(g + h))(c2g(h + f) – b2h(f + g)),
(5.2)
and y, z may be obtained from Equation (5.2) by cyclic change of a, b, c and f, g, h. It may now be checked that Y lies on the circumcircle of triangle EFG. Similarly Y lies on the nine-point circles of triangles ABD, ACD and BCD. 6. L, M, N the midpoints of the sides FG, GE, EF L is the midpoint of FG and therefore has co-ordinates (f(g + h), g(f + 2g + h), h(f + g)). The points M, N have co-ordinates that may be written down from those of L by cyclic change of x, y, z and f, g, h. It may now be checked that L, U, X, T are collinear with UX = XT. Similarly M, P, X, R are collinear with PX = XR and N, S, X, Q are collinear with SX = XQ. The point Z on YX such that YX = XZ also lies on the hyperbola Σ (since X is the centre) and if ABCD is cyclic Z becomes the circumcentre of ABCD.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2 SP.
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Article: CJB/2011/145
Concyclic Circumcentres in the Steiner Configuration Christopher Bradley
A
E
N
I P
F
T
U
M G
J
R
Q S B
L
K
C
D
Fig. 1 The Steiner Ellipses and the Circle of Circumcentres Abstract: In triangle ABC with centroid G, the points D, E, F lie on AG, BG, CG respectively and are such that AG = GD, BG = GE, CG = GF. It is proved that the circumcentres of triangles AFG, FBG, BGD, DGC, CGE, EGA are concyclic. 1. Summary of conclusions The six triangles are congruent with sides equal to 2/3 of the medians. In proving this, note the existence of parallelograms such as AFBG. They therefore have the same circumradius. It follows they lie on a circle centre G. 2. Results of computations The ellipse PQRSTU is the inner Steiner circle with equation x2 + y2 + z2 – 2yz – 2zx – 2xy = 0.
1
(2.1)
The radius r of the circumcircle of circumcentres is given by r2 = (2b2 + 2c2 – a2)(2c2 + 2a2 – b2)(2a2 + 2b2 – c2)/(81(a + b + c)(b + c – a)(c + a – b)(a + b – c)) (2.2)
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/2011/146
More Conics in a Semi-Regular Hexagon Christopher Bradley Abstract: ABC is an equilateral triangle and AFBDCE is a hexagon inscribed in a conic. Triangles ABC and DEF are in perspective with vertex P, the centroid of ABC. G1- G6 are the centroids of triangles EAF, FBD, DCE, BDC, CEA, AFB. It is shown that these points lie on a conic. Also triangles G1G2G3 and G4G5G6 are congruent and in perspective. The six interior points of intersection of triangles ABC and DEF also lie on a conic.
A
G1 E 21 F
G5
31
G6
32
Q
P
23
G3 G2
13
12
C
B G4
D Fig. 1 1. Points D, E, F
1
We use areal co-ordinates with ABC as triangle of reference. Since P is the centroid of ABC its co-ordinates are (1, 1, 1). The equation of AP is y = z and this meets the conic with equation fyz + gzx + hxy = 0 (1.1) at the point D with co-ordinates D(– f, g + h, g + h). Similarly E, F have co-ordinates E(h + f, – g, h + f) and F(f + g, f + g, – h). 2. The Centroids G1 – G6 The co-ordinates of a centroid have co-ordinates which are equal to 1/3 of the normalized coordinates of its vertices. So in triangle AFE, the point F must be written with normalized coordinates (1/(2f + 2g + h))(f + g, f + g, – h). This means that the co-ordinates of the centroids are technically more involved than one may initially anticipate. The co-ordinates of G1, the centroid of triangle AFE, are (x, y, z), where x = (1/k)(8f2 + 5f(g + h) – 3(g2 – 3gh + h2)), y = (1/k)(2f2 + f(2h – g) – 3g(g – h)), z = (1/k)(2f2 + f(2g – h) + 3h(g – h)).
(2.1) (2.2) (2.3)
Here k = 3(2f + 2g – h)(2h + 2f – g).
(2.4)
The co-ordinates of G2 and G3 follow from those of G1 by cyclic change of x, y, z and f, g, h. The co-ordinates of G6, the centroid of triangle AFB, are (x, y, z), where x = (1/j)(3f + 3g – h), y = (1/j)(3f + 3g – h), z = (1/j)(– h).
(2.5) (2.6) (2.7)
Here j = 3(2f + 2g – h).
(2.8)
The co-ordinates of G4, G5 follow from those of G6 by cyclic change of x, y, z and f, g, h. 3. The Conic passing through the six Centroids It may now be checked that the six centroids lie on the conic with equation of the form ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, where
(3.1)
u = 2(f – 3g – 3h)(2f2 + f(g + h) – (g – h)2)
(3.2)
2
with v, w following from u by cyclic change of f, g, h and l = 2f3 + 14f2(g + h) + f(7g2 + 16gh + 7h2) – 5g3 + 9g2h + 9gh2 – 5h3,
(3.3)
with m, n following from l by cyclic change of f, g, h. 4. The Points of Intersection of the Triangles ABC and DEF Points of intersection are labelled as follows: AB^EF = 31, AB ^FD = 32, BC^ FD = 12, BC^DE = 13 CA^DE = 23, CA^EF = 21. The equations of EF, FD, DE are: fx – (h + f)y – (f + g)z = 0, gy – (f + g)z – (g + h)x = 0, hz – (g + h)x – (h + f)y = 0.
(4.1) (4.2) (4.3)
The co-ordinates of the points of intersection now follow and are: 31: (h + f, f, 0); 21(f + g, 0, f); 12(0, f + g, g); 32(g, g + h, 0); 23(h, 0, g + h); 13(0, h, h + f). (4.4) 5. The Conic containing the six Points of Intersection It may now be checked that the equation of the conic passing through these six points of intersection is of the form (3.1) with u = 2f(g + h), v = 2g(h + f), w = 2h(f + g), l = – f2 – f(g + h) – 2gh, m = – g2 – g(h + f) – 2hf, n = – h2 – h(f + g) – 2fg. (5.1)
6. The Triangles G1G2G3 and G4G5G6 It is a short calculation to show that these two triangles are congruent with their sides parallel. They are also in perspective at the point Q with co-ordinates (x, y, z), where x = 4f3 – 3f2(g + h) – f(5g2 + 2gh + 5h2) + 2(g3 – 2g2h – 2gh2 + h3) , (6.1) with y, z following from x by cyclic change of f, g, h. Cabri also indicates the finite points of intersection lie on a conic, and this follows from the fact that the three other points of intersection lie on a line (the line at infinity) – indeed two triangles must either intersect on a line and a conic or on a cubic curve. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article: CJB/2011/147
The Midpoint Rectangular Hyperbola again Christopher Bradley Abstract: If A, B, C, D are four points on a circle and E is any other point, then the locus of the centre of conic ABCDE as E varies is the rectangular hyperbola through the midpoints of the quadrangle ABCD.
A
O G
E
Q B
P
D
T
Fig. 1 The Midpoint Rectangular Hyperbola
1
C
1. Introduction This famous result (see the Abstract) is verified in the simple case in which ABC is equilateral and D is a point on circle ABC with simple areal co-ordinates, which are used throughout with ABC as triangle of reference. In Fig. 1 ABC is an equilateral triangle, so we may take a = b = c = 1. The equation of the circle ABC is yz + zx + xy = 0 (1.1) We take D to have the simple co-ordinates D(– 1, 2, 2). Now we choose a point E(m , n, 1 – m – n) and consider the conic ABCDE. Suppose its equation is fyz + gzx + hxy = 0.
(1.2)
Then, since it passes through D, we have 2f = g + h,
(1.3)
and, since it passes through E, we have fn(m + n – 1) + m(g(m + n – 1) – hn) = 0.
(1.4)
2. The conic ABCDE and its centre Solving equations (1.3) and (1.4) we obtain f = m(m + 2n – 1), g = n(m – n + 1), h = (m + n – 1)(2m + n),
(2.1)
which values when substituted back in Equation(1.2) gives the equation of the comic ABCDE. The centre of the conic has co-ordinates(x, y, z), where x = – f2 +fg + hf = m2(m + 2n – 1), y = – g2 + gh + fg = n(m – n + 1)(3m2 + m(4n – 3) + 2n(n – 1)), z = – h2 + hf + gh = – (m + n – 1)(m2 – m + 2n(n – 1))(2m + n).
(2.2) (2.3) (2.4)
3. The locus of the centre It may now be checked that the locus of the centre O of this conic has equation 2x2 – y2- z2 + 2yz – zx – xy = 0.
2
(2.5)
This curve obviously passes through the centre G of triangle ABC and the centres of the sides of the cyclic quadrilateral ABCD, and hence is the rectangular hyperbola (see Article 144). Its centre is the point Q(2, 5, 5). (
3
Article: CJB/2011/148
A Circle through two Vertices, three Circumcentres and a Miquel Point Christopher Bradley Abstract: Starting with a triangle ABC and a point D on BC it is shown how the circumcentres of triangles ABC, ADB, ADC lie on a circle through A. If this circle intersects circle ABC at A and P, then it is shown that triangle PBC exhibits a Miquel point Q with points D, M, N on its sides.
A P
M
N O Q B
D
C
Fig. 1 A circle with two vertices, three circumcentres and a Miquel point 1. Introduction
1
Cartesian co-ordinates are used in dealing with the construction in Fig. 1. A triangle ABC is given and D is a point on BC, We take A as origin and BC with equation x = 1, B, D, C having y = – v, u, w respectively. O, N, M are the circumcentres of triangles ABC, ABD, ACD and it is proved that circle OMN passes through A. (The situation is similar to the configuration of circles of a quadrangle, but although simpler, it is not a result that I have seen before.) Circle OMNA meets circle ABC again at a point P and surprisingly N lies on BP and M lies on CP. When this is proved then the triangle PBC has D, M, N on its sides, so that the Miquel point Q of D, M, N on the sides of triangle PBC, also lies on circle OMNAP. In the following sections we provide proofs and obtain co-ordinates of O, M, N, P and Q. 2. The circles ABC, ABD, ACD and their centres O, N, M The equation of circle ABC is x2 + y2 – (1 + vw)x – (w – v)y = 0,
(2.1)
and its centre O has co-ordinates O((1 + vw)/2, (w – v)/2). The equation of circle ABD is x2 + y2 – (1 + uv)x – (u – v)y = 0,
(2.2)
and its centre N has co-ordinates N((1 + uv)/2, (u – v)/2). The equation of circle ACD is x2 + y2 – (1 – uw)x – (u + w)y = 0,
(2.3)
and its centre M has co-ordinates M((1 – uw)/2, (u + w)/2). 3. The circle OMN passes through A and circles OMNA and ABC meet again at P The equation of the circle OMN is 2x2 + 2y2 – (1 + vw – wu + uv)x – (uvw + u – v + w)y = 0.
(3.1)
This evidently passes through A(0, 0). Circles OMN and ABC with equations (2.1) and (3.1) meet at A and P (x, y), where x = u(uvw + u + v – w)/(1 + u2), y = u(1 + vw + wu – uv)/(1 + u2). 2
(3.2) (3.3)
4. The lines BN and CM pass through P The equation of BN is (u + v)x + (1 – uv)y = u(1 + v2).
(4.1)
It may now be checked that BN passes through P. Similarly CM passes through P. 5. Circles OMNAP, CMD and BND meet at the Miquel point Q Circle OMNAP has equation (2.1). Circle CMD has equation 2(1 + wu)(x2 + y2) + (w2u2 – u2 – w2 – 3)x + 2(w + u)(1 + wu)y + (1 + u2) (1 + w2) = 0.
(5.1)
The Miquel point Q is the intersection of circles with equations (3.1) and (5.1) and has coordinates (x, y), where x = (1/k)(u2(1 – v2w2) + v2 + w2 + 2), (5.2) 2 2 2 y = (1/k)(u ((vw – 1)(v – w)) + u(1 + v ) (1 + w ) + vw(v – w) + v + w), (5.3) where k = u2(1 – vw)2 + 2u(vw – 1)(v – w) + (v – w)2 + 4.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
(5.4)
Article: CJB/2011/149
Circles with a common point in a Cyclic Quadrilateral Christopher Bradley Abstract: If ABCD is a cyclic quadrilateral it is possible to find points P, Q, R, S, T, U on sides AB, BC, CD, DA, AC, BD respectively so that circles BPQU, APST, CQRT, DRSU have a common point M. These circles have centres that are concyclic, as are points M, U, I, T where I = AC^BD.
B P F A
M
E K
U
S
H
Q
I
T
G D R
C
Fig. 1 A Common Miquel point M for triangles ABD, CBD, ABC, ADC in a cyclic quadrilateral 1. The construction 1
The important thing is that the above figure does not merely exist but can be constructed. There are, in fact, three degrees of freedom, which may be thought of as allowing free positioning of points P, Q and S. The construction then is as follows: Choose P and Q and draw circle BPQ to meet BD at U. Choose S and draw circle SDU to meet CD at R and circle BPQU again at M. Then draw circle CRQ and discover that it passes through M and meets AC at T. Now draw circle APS and discover that it passes through M and T. Let AC and BD meet at I and discover that I, T, M, U are concyclic. Finally if E, F, G, H are the centres of circles APS, BPQ, CQR, DRS respectively then discover that E, F, G, H are concyclic. Note that the results other than the concyclic properties (EFGH, ITMU) are true for any quadrilateral. The concyclic properties are true if ABCD is a cyclic quadrilateral. In the following sections the above properties are proved analytically using areal co-ordinates. It is all technically rather difficult and so certain results are left for the interested reader to check (using algebraic software package such as DERIVE, which we used). 2. Circle BPQ and the point U We take ABC as triangle of reference and as we are first considering any quadrilateral we take D to have co-ordinates D(f, g, h). We also take P on AB to have co-ordinates (p, 1 – p, 0) and Q on BC to have co-ordinates Q(0, q, 1 – q). Eventually we shall suppose D lies on circle ABC and then we set f = – a2t(1 – t), g = b2(1 – t), h = c2t, where t is a parameter, – ∞ < t < = ∞. The circle BPQ has an equation of the form a2yz + b2zx + c2xy + (lx + my + nz)(x + y + z) = 0,
(2.1)
where we find l = – c2(1 – p), m = 0, n = – a2q.
(2.2)
The diagonal BD, with equation hx = fz, meets circle BPQ at the point U with co-ordinates (x, y, z), where x = f(c2fp + a2h(1 – q)), y = a2hq(f + h) – f(b2h – c2(f + h)(1 – p)), z = h(c2fp + a2h(1 – q)). (2.3) 3. Circle SDU and the points R and M The equation of AD is hy = gz, so we may take S to have co-ordinates S(s, g, h). Circle SDU now has an equation of the form (2.1), where we find l = (a2gh – (g + h)(b2h + c2g))/((f + g + h)(g + h + s)), m = (a2h(q – 1) – c2fp)/(f + g + h), 2
(3.1) (3.2)
n = – (a2gh(f + q(g + h + s) + b2fhs – c2fg(gp + hp +s(p – 1))/(h(f + g + h)(g + h + s)).
(3.3)
The equation of CD is gx = fy and this meets circle SDU at the point R with co-ordinates (x, y, z), where x = f(a2gh(f + q(g + h + s)) + f(b2hs – c2g(gp + hp + s(p – 1)))), (3.4) 2 2 2 y = g(a gh(f + q(g + h + s)) + f(b hs – c g(gp + hp + s(p – 1)))), (3.5) 2 2 2 z = f(b h(f + g)(g + h) + c g(f(gp + hp + s(p – 1)) + g p + g(h + s)(p – 1) – h(h + s))) – a2gh(f2 + f(gq + (h + s)(q – 1)) + g(g + h + s)(q – 1)). (3.6) The equation of the common chord of two circles with equations of the form (2.1) with l1, m1, n1 and l2, m2, n2 respectively is (l1 – l2)x + (m1 – m2)y + (n1 – n2)z = 0. (3.7) Circles BPQ and SDU meet at U and a second point M and from Equations (2.2) and (3.1) – (3.3) we find the equation of the line UM is h(a2gh – b2h(g + h) – c2(f(g + h + s)(p – 1) + (g + h)(gp + (h + s)(p – 1))))x + h(g + h + s)(a2h(q – 1) – c2fp)y + a2h(f(g(q – 1) + q(h + s) + hq(g + h + s)) – f(b2hs – c2g(gp + hp + s(p – 1)))z = 0. (3.8) This line meets circle BPQ at U (whose co-ordinates are given by Equations (2.3) and the point M with co-ordinates (x, y, z), where x = a2h(g + h + s)(a2h(q – 1)(g(q – 1) + q(h + s)) + b2hs(1 – q) + c2(g2p(q – 1) + g(hp(2q – 1) + s(p – 1)(q – 1)) + hpq(h + s))), (3.9) y = – a4gh2(g(q – 1) + q(h + s)) + a2h(b2h(g2(q – 1) +g(h(2q – 1) + s(q + 1)) + hq(h + s)) + c2g(g2p(q – 2) + g(h(p(2q – 3) + 1) + s(2p(q – 1) – q + 2)) + h2(p(q – 1) + 1) + hs(p(2q – 1) – q + 1) + qs2(p – 1))) – (g + h)(b4h2s – b2c2h(g2p + g(hp – s) + s(1 – p)(h + s)) – c4g(g2p2 + gp(h(2p – 1) + 2s(p – 1)) + (p – 1)(h2p + hs(2p – 1) + s2(p – 1)))), (3.10) 2 2 z = c h(g + h + s)(a h(g(p(q – 2) – q + 1) + h(p – 1)(q – 1) + s(p – 1)(q – 1)) + p(g + h)(b2h + c2(gp + h(p – 1) + s(p – 1)))). (3.11) 4. Circles CRQ and APS pass through M and the points T and I Circle CRQ has the form of Equation (2.1) and we find l = (a2gh – b2h(g + h) – c2g(gp + hp + s(p – 1)))/(h(g + h + s)), m = a2(q – 1), n = 0. It may now be checked that circle CRQ passes through M. Circle APS has the form of Equation (2.1) and we find 3
(4.1) (4.2) (4.3)
l = 0, m = – c2p, n = – (a2gh + b2hs – c2g(gp + hp + s(p – 1)))/(h(g + h + s)).
(4.4) (4.5) (4.6)
It may now be checked that circle APS passes through M. Hence M is a common point of all four circles. The circle APS meets AC with equation y = 0 at A and the point T with co-ordinates (x, y, z), where x = – a2gh – b2hs + c2g(gp + hp + s(p – 1)), (4.7) y = 0, (4.8) 2 2 2 z = a gh – b h(g + h) – c g(gp + hp + s(p – 1)). (4.9) It may now be checked that circle CRQ passes through T. The diagonals AC and BD meet at I(f, 0, h). It has now been established that if ABCD is any quadrilateral and arbitrary points P, Q, S are chosen on AB. BC, DA then points R, T, U exist on CD, CA, BD respectively and a point M exists so that circle BPQ passes through M and U, circle DSU passes through R and M, circle CRQ passes through T and M and circle APS passes through T and M. Thus M is a common Miquel point of the four triangles ABD, BCD, ACD and ABC. 5. The centres E, F, G, H of the circles APS, BPQ, CRQ, DRS The centre of a circle whose equation is of the form (2.1) has co-ordinates (x, y, z) given by the equations x = – (a4 – a2(b2 + c2 + 2l – m – n) + (b2 – c2)(m – n)), (5.1) 4 2 2 2 2 2 y = – (b – b (c + a + 2m – n – l) + (c – a )(n – l)), (5.2) 4 2 2 2 2 2 z = – (c – c (a + b + 2n – l – m) + (a – b )(l – m)). (5.3) Using the values of l, m, n given by Equations (2.2) we find the co-ordinates of F, the centre of circle BPQ to be (x, y, z), where x = a2(a2(q – 1) + b2(1 – q) + c2(2p + q – 1)), (5.4) 4 2 2 2 4 2 2 4 y = – a q + a (b (q + 1) – c (p – q – 1)) – b + b c (2 – p) + c (p – 1), (5.5) 2 2 2 2 z = c (p(b – c ) – a (p + 2q – 2)). (5.6) Using the values of l, m, n given by Equations (3.1) – (3.3) we find the co-ordinates of H, the centre of circle DRS to be (x, y, z), where
4
x = – (1/(h(f + g + h)(g + h + s)))(a4h(f(h + s) + g2(1 – q) – g(h + s(q – 1)) + hq(h + s)) – a2(b2h(f(h + 2s) + g2(1 – q) – g(h(2q – 1) + s(q – 1)) – h(hq + s(q – 2))) + c2(h(g2(q – 1) + g(h(2q – 1) + s(q + 1)) + hq(h + s)) – f(g2p + g(s(p – 1) – 2h) – h(h + s)(p + 1)))) + f(b2 – c2)(b2hs – c2(g2p + g(2hp + s(p – 1)) + hp(h + s)))), (5.7) y = – (1/(h(f + g + h)(g + h + s)))(a4gh(f + gq + h(q + 1) + qs) – a2( b2h(f(2g + h) + g2(q + 1) + g(3hq + s(q + 1)) + h(2hq + s(2q – 1))) + c2g(f(gp + h(p + 1) + s(p – 1)) + h(g(q + 1) + h(q + 2) + qs))) + b4h(f + g + s)(g + h) + b2c2(f(g2p + g(h(3p – 1) + s(p – 1)) + h(h(2p – 1) + 2s(p – 1))) – h(g + h)(2g + s)) + c4g(f(gp + hp + s(p – 1)) + h(g + h))). (5.8) 4 2 2 2 z = (1/(h(f + g + h)(g + h + s))(a h (g(q – 2) + (h + s)(q – 1)) – a h(b h(g(q – 3) + h(q – 2) + s(q – 1)) + c2(f(g(p + 1) + (h + s)(p – 1)) + 2g2(q – 1) + g(3h(q – 1) + s(2q – 1)) + h(h + s)(q – 2))) – b4h2(g + h) + b2c2h(f(g(p + 1) + h(p + 1) + s(p – 1)) + (g + h)(2h + s)) + c4(f(2g2p + g(h(3p – 1) + 2s(p – 1)) + h(h + s)(p – 1)) – h(g + h)(h + s))). (5.9) Using values of l, m, n given by Equations (4.1) – (4.3) we find the co-ordinates of G, the centre of circle CRQ are (x, y, z), where x = – (1/(h(g + h + s)))(a2(a2h(g(q – 2) + q(h + s)) + b2h(gq + hq + s(q – 2)) + c2(2g2p + g(h(2p – q) + 2s(p – 1)) – hq(h + s)))), (5.10) 4 2 2 2 y = – (1/(h(g + h + s)))(a gh – a (b h(g(2q – 1) + 2hq + s(2q – 1)) + c g(gp + h(p + 1) + s(p – 1))) + b4hs – b2c2(g2p + g(hp + s(p – 1)) + hs) + c4g(gp + hp + s(p – 1))), (5.11) 4 2 2 2 z = (1/(h(g + h + s)))(a h(g(q – 2) + (h + s)(q – 1)) + a (c (g p + g(h(p – q + 1) + s(p – 1)) + h(2 – q)(h + s)) – b2h(g(q – 3) + h(q – 2) + s(q – 1))) – b4h(g + h) – b2c2(g2p + g(h(p – 2) + s(p – 1)) – h(2h + s)) + c4(g2p + g(h + s)(p – 1) – h(h + s))). (5.12) Using values of l, m, n given by Equations (4.4) – (4.6) we find the co-ordinates of E, the centre of circle APS are (x, y, z), where x = – (1/(h(g + h + s)))(a4h(h + s) + a2(c2(g2p + g(s(p – 1) – 2h) – h(h + s)(p + 1)) – b2h(h + 2s)) + (b2 – c2)(b2hs – c2(g2p + g(2hp + s(p – 1)) + hp(h + s)))), (5.13) 4 2 2 2 4 y = – (1/(h(g + h + s)))(a gh – a (b h(2g + h) + c g(gp + h(p + 1) + s(p – 1))) + b h(g + h) + b2c2(g2p + g(h(3p – 1) + s(p – 1)) + h(h(2p – 1) + 2s(p – 1))) + c4g(gp + hp + s(p – 1))) (5.14) 2 2 2 z = – (1/(h(g + h + s)))(c (a h(g(p + 1) + (h + s)(p – 1)) – b h(g(p + 1) + h(p + 1) + s(p – 1)) – c2(2g2p + g(h(3p – 1) + 2s(p – 1)) + h(h + s)(p – 1)))). (5.15) Using f = – a2t(1 – t), g = b2(1 – t), h = c2t,
(5.16)
so that D lies on circle ABC it may now be shown that E, F, G, H are concyclic. In general they are not concyclic. The working is technically very difficult and lengthy (covering about 22 pages
5
of detailed working in DERIVE, which cannot be relied upon to be typed accurately – an unfortunate property of DERIVE is that one cannot copy from it into a Word text). 6. The points I, T, M, U are concyclic when ABCD is a cyclic quadrilateral The co-ordinates of U are given by Equations (2.3) and using Equations (5.16) these become x = a4c2t2(t – 1)(p(t – 1) – q + 1) (6.1) 2 2 2 2 2 2 y = – a c t (a (t – 1)(p(t – 1) – q – t + 1) + b (t – 1) + c (p(t – 1) – q – t + 1)), (6.2) 2 4 2 z = a c t (p(t – 1) – q + 1). (6.3) The co-ordinates of M are given by Equations (3.9) – (3.11) and using Equations (5.16) these become x = a2c4t(a2t(q – 1)(b2(q – 1)(t – 1) – q(c2t + s)) + b4p(1 – q)(t – 1)2 + b2(c2pt(t – 1)(2q – 1) + s(q – 1)(p(t – 1) + 1)) – c2pqt(c2t + s))(b2(t – 1) – c2t – s), (6.4) 4 2 4 2 2 2 2 2 4 4 2 y = a b c t (1 – t)(b (q – 1)(t – 1) – q(c t + s)) – a b c t(b (t – 1) (p(q – 2)(t – 1) – t(q – 1)) + b2(1 – t)(c2t(p(t – 1)(2q – 3) – 2qt + 2t – 1) + s(2p(q – 1)(t – 1) + q(1 – 2t) + t – 2)) + c4t2(p(q – 1)(t – 1) – qt + t – 1) + c2st(p(t – 1)(2q – 1) + q(1 – 2t) + t – 1) + qs2(p – 1)(t – 1)) + b2c4(b2(t – 1) – c2t)(b4p(t – 1)2(p(t – 1) – t) – b2(c2pt(t – 1)(2p(t – 1) – 2t + 1) + s(2p2(t – 1)2 – 2p(t – 1)2– t)) + (p – 1)(c4pt2(t – 1) + c2st(2p(t – 1) + 1) + s2(p(t – 1) + 1))), (6.5) 6 2 2 2 z = c t(a t(b (t – 1)(p(q – 2) – q + 1) + c t(1 – q)(p – 1) + s(1 – q)(p – 1)) + p(b2(p(t – 1) – t) + (1 – p)(c2t+ s))(c2t – b2(t – 1)))(b2(t – 1) – c2t – s). (6.6) The co-ordinates of I given in Section 4 become x = – a2t(1 – t), y = 0, z = c2t.
(6.7)
The co-ordinates of T given by Equations (4.7) – (4.9) become on using Equation (5.16) x = a2b2c2t(t – 1) + b2c2(b2p(t – 1)2 + c2pt(1 – t) – s(p(t – 1) + 1)), (6.8) y = 0, (6.9) 2 2 2 2 2 2 2 z = a b c t(1 – t) – b c (b (t – 1)(p(t – 1) – t) + c t(t – p(t – 1)) + s(1 – t)(p – 1)). (6.10) It may now be shown that U, M, I, T lie on a circle with equation of the form (2.1) with l = b2c2(a2t(t – 1) + b2(t – 1)(p(t – 1) – t) + c2t(t – p(t – 1)) + s(1 – t)(p – 1)) all divided by t(a2(t – 1) + c2)(b2(t – 1) – c2t – s), m = – (a2c2(p(t – 1) – q + 1))/(a2(t – 1) + c2), n = a2b2(1 – t)(a2t(t – 1) + b2p(t – 1)2 + c2pt(1 – t) + ps(1 – t) – s), all divided by t(a2(t – 1) + c2)(b2(t – 1) – c2t – s).
(6.11) (6.12) (6.13)
The points I, T, M, U are concyclic whether or not ABCD is a circle. David Monk has kindly pointed out that EG is the line of centres of circles 6
APS and CQR and therefore bisects the common chord MT at right angles. Similarly FH bisects MU at right angles and hence the chords EG, FH intersect at K, the centre of circle ITMU. Fig. 2 shows the construction when the quadrilateral is re-entrant.
A T E S P
M
K G
H I
R
U D
F
B
C
Q
Fig. 2 Showing the common point M when ABCD is re-entrant Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL, BS8 2SP.
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Article: CJB/2011/150
When a Point is a Circumcentre, an Incentre and an Orthocentre Christopher Bradley Abstract: Given a triangle ABC, circumcentre O, and the centres D, E, F of circles BOC, COA, AOB respectively, then O is the incentre of triangle DEF, as well as being the orthocentre of the triangle of excentres. W' A'
F'
To A'
To F'
To W' A
F
U
E' E To E'
J
O'
O
V'
To V' P
Q W
V
C
B
To U' To B'
To C' C'
D
B'
To D'
D' Fig. 1 Showing O as a circumcentre of triangle ABC and incentre of triangle DEF
1
U'
1. Introduction The above figure is not as complicated as appears at first sight. Given a triangle ABC with circumcentre O, it shows circles BOC, COA, AOB with centres D, E, F respectively. It is shown that O is the incentre of DEF touching its sides at U, V, W. Also shown is that AD, BD, CD are concurrent at a point P. Points A', B', C' are the excentres of triangle DEF and OD, OE, OF are perpendicular to the sides of triangle A'B'C'. Circle DEF has centre J and circle A'B'C' has centre O' and it is proved that O, J, O' are collinear. Also shown is that AA', BB', CC' are concurrent at a point Q on the line of centres. Line OD passes through one point of intersection of circles BOC and A'B'C', the other point lying on AA', and similarly for lines OE, OF. We do not prove, but CABRI indicates, that if X is the second point of intersection of circle DEF and B'C' and with similar definition of Y, Z, then AX, BY. CZ are concurrent at a point on the line of centres. In the following sections we prove these assertions, using areal co-ordinates with ABC as triangle of reference. 2. Circles BOC, COA, AOB and their centres D, E, F With ABC as triangle of reference the circumcentre O has co-ordinates (a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)).
(2.1)
The circle BOC has an equation of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0,
(2.2)
where u, v, w are determined by the co-ordinates of B, O, C. The result is that v = w = 0 and u = b2c2/(a2 – b2 – c2), (2.3) so that its equation is b2c2x2 – a2yz(b2 + c2 – a2) + b2(a2 – b2)zx + c2(a2 – c2)xy = 0.
(2.4)
Circle COA has w = u = 0 and v = c2a2/(b2 – c2 – a2)
(2.5)
and its equation may be obtained from Equation (2.4) by cyclic change of x, y, z and a, b, c. Similarly circle AOB has u = v = 0 and w = a2b2/(c2 – a2 – b2) and its equation may be obtained by another cyclic change of x, y, z and a, b, c. The centre of a circle of the form (2.1) has co-ordinates x, y, z, where 2
(2.6)
x = – (a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w)), y = – (b4 – b2(c2 + a2 + 2v – w – u) + (c2 – a2)(w – u)), z = – (c4 – c2(a2 + b2 + 2w – u – v) + (a2 – b2)(u – v)). Using Equation (2.3), we find the co-ordinates of D are (x, y, z), where x = – a2(a4 + b4 + c4 – 2a2(b2 + c2)), y = b2(a4 + b4 – 2a2b2 – b2c2 – c2a2), z = c2(c4 + a4 – 2a2c2 – a2b2 – b2c2).
(2.7) (2.8) (2.9)
(2.10) (2.11) (2.12)
The co-ordinates of E and F follow from Equations (2.10) – (2.12) by cyclic change of x, y, z and a, b, c. For example, the x-co-ordinate of E is a2(a4 + b4 – 2a2b2 – b2c2 – c2a2). 3. Lines AD, BE, CF meet at a point P The equation of the line AD is c2(c4 + a4 – 2a2c2 – a2b2 – b2c2)y = b2(a4 + b4 – 2a2b2 – b2c2 – c2a2)z.
(3.1)
The equations of BE and CF follow from Equation (3.1) by cyclic change of x, y, z and a, b, c. It may now be shown that these three lines are concurrent at a point P with co-ordinates (x, y, z), where x = a2/((b2 – c2)2 – a2(b2 + c2)), (3.2) 2 2 2 2 2 2 2 y = b /((c – a ) – b (c + a )), (3.3) z = c2/((a2 – b2)2 – c2(a2 + b2)). (3.4) 4. The triangle DEF and its incircle UVW centre O The equation of EF is a2b2c2x + c2(a4 + b4 + c4 – a2b2 – 2b2c2 – 2c2a2)y + b2(a4 + b4 + c4 – a2c2 – 2b2c2 – 2a2b2)z = 0. (4.1) The line EF meets the line OA at a point U with co-ordinates (x, y, z), where x = – 2a4 – b4 – c4 + 2b2c2 + 3a2(b2 + c2), y = b2(c2 + a2 – b2), z = c2(a2 + b2 – c2).
(4.2) (4.3) (4.4)
V = FD^OB, W = DE^OC have co-ordinates that may be written down from those of U by cyclic change of x, y, z and a, b, c. Circle UVW has an equation of the form (2.2) with 3
u = v = w = 3a2b2c2/(4(a4 + b4 + c4 – 2b2c2 – 2c2a2 – 2a2b2)).
(4.5)
It may now be shown that EF touches circle UVW and similarly for FD, DE and so UVW is the incircle of triangle DEF. Using Equations (2.7) to (2.9) we find the circle UVW has centre O. 5. Points D', E', F', A', B', C' and circle D'E'F' OD meets circle BOC at the point D' whose co-ordinates turn out to be D'(– a2, b2, c2) and similarly for E' and F'. Thus D', E', F' are the ex-symmedians of triangle ABC. They lie on a circle with equation of the form (2.2), where u = a2b2c2/((a2 + b2 – c2)(c2 + a2 – b2)). (5.1) v and w follow by cyclic change of a, b, c. The line OD meets circle D'E'F' at D' and again at a point A' with co-ordinates (x, y, z), where x = a2(a6 – a4(b2 + c2) – a2(b2 + c2)2 + (b2 + c2)(b2 – c2)2, (5.2) 2 6 4 2 2 2 4 4 2 2 3 y = – b (a – a (b + c ) – a (b – c ) + (b – c ) ), (5.3) 2 6 4 2 2 2 4 4 2 2 3 z = c (a – a (b + c ) + a (b – c ) – (b – c ) ). (5.4) A', B', C' are the excentres of triangle DEF. 6. Circle DEF, centre J and circle D'E'F', centre O' and O, J, O' collinear Circle DEF has an equation of the form (2.2) with u = a4b2c2(a4 + b4 + c4 + b2c2 – 2a2(b2 + c2)) all divided by (b2 + c2 – a2)(c2 + a2 – b2)(a2 + b2 – c2)(2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4),
(6.1)
and v, w follow from u by cyclic change of a, b, c. Using Equation (6.1) we find the co-ordinates of the centre J of circle DEF has co-ordinates (x, y, z), where x = a2(a8 – 2a6(b2 + c2) + 2a4b2c2 + a2(2b6 + 2c6 – b2c2(b2 + c2)) – (b6 – c6)(b2 – c2) (6.2) 2 2 2 2 2 2 2 2 2 all divided by (b + c – a )(c + a – b )(a + b – c ). y and z follow from x by cyclic change of a, b, c.
4
From Equation (5.1) the centre O' of circle D'E'F' has co-ordinates (x, y, z), where x = a2(a8 – 2a6(b2 + c2) + 2a2(b6 + c6) + (b6 – c6 – b2c2(b2 – c2))(c2 – b2))
(6.3)
all divided by (b2 + c2 – a2)(c2 + a2 – b2)(a2 + b2 – c2). Now y and z follow from x by cyclic change of a, b, c. It may now be checked that points O, J, O' are collinear. 7. AA', BB', CC' are concurrent at a point Q on the line of centres Line CC' has equation b2(a6 – a4(3b2 + c2) + a2(3b4 + c4) – b6 + b4c2 + b2c4 – c6)x + a2(a6 – a4(3b2 + c2) + a2(3b4 – c4) – b6 + b4c2 – b2c4 + c6)y
(7.1)
Lines AA' and BB' have equations that may be written down from Equation (7.1) by cyclic change of x, y, z and a, b, c. It may now be checked that these three lines are concurrent at a point Q with co-ordinates (x, y, z), where x = a2(c2 + a-2 – b2)(a2 + b2 – c2)(2a2(b2 + c2) – a4 – b4 – c4) (7.2) and y, z follow by cyclic change of a, b, c. It may now be checked that Q is collinear with O, J, O'. 8. The points U', V', W' lie on circles BOC, COA, AOB respectively We define U' to be the second point of intersection of AA' and circle A'B'C'. It has co-ordinates (x, y, z), where x = a2(c2 + a2 – b2)(a2 + b2 – c2)(b2 + c2 – a2), y = b2(a6 – a4(b2 + c2) + a2(c4 – b4) + (b2 – c2)3), z = c2(a6 – a4(b2 + c2) + a2(b4 – c4) – (b2 – c2)3).
(8.1) (8.2) (8.3)
It may now be checked that U' lies on circle BOC. Similarly V', W' lie on circles COA, AOB respectively. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 5
Article: CJB/2011/151
The Orthocentre Conics Christopher Bradley Abstract: When circles BHC, CHA, AHB in a triangle ABC with orthocentre H are drawn their centres D, E, F have interesting properties.
ef
ca
A
ab
21
31
F
E
L
32 O N
H
23
M B
13
12 D
C de
fd
bc
Fig. 1 The three orthocentre conics Brief description This is a very familiar configuration so that all we choose to do is to give a brief description of the results. ABC is a triangle, with orthocentre H and D, E, F are the centres of the circles BHC, CHA, AHB respectively. Triangle ABC and DEF are congruent and map into one another by a 180o rotation about the nine-point centre N. Points A, B, C, D, E, F lie on a conic, centre N 1
The six finite intersections of triangles ABC and DEF (points 12, 13, 23, 21, 31, 32 in the figure) also lie on a conic centre N. Points 12 21 N are collinear with 12 N = N 12 and similarly for other pairs of points 23 32 and 31 13. The tangents to conic ABCDEF at it vertices meet in pairs ab, bc, ca, de, ef, fd (with ab the intersection of tangents at A and B etc.) These six points also lie on a conic, centre N and once again point pairs such as ab, de lie on a diameter.
2
Article: CJB/2011/152
Generating Circles from the Symmedian point Christopher Bradley Abstract: Given a triangle ABC and its symmedian point K circles BKC, CKA, AKB are drawn. Their other points of intersection with the sides of ABC are shown to lie on a Tucker circle. These points also lie in pairs on the sides of another triangle and the other intersections of these sides with the circles BKC, CKA, AKB are also concyclic. Centres of these circles are shown to lie on the Brocard axis.
W2
A
W
12
V1
F
W1
13
V
E V3 O
K
21
B
P
Q C
23
D
31
U3
U2
32
U
Fig. 1 1
1. Introduction The construction of Fig. 1 is reasonably straightforward. Given triangle ABC, symmedian point K and circumcentre O, the circles BKC, CKA, AKB centres D, E, F respectively are drawn. The second intersections of circle BKC with AB and AC are 21 and 31 respectively. The second intersections of circle CKA with AB and BC are 12 and 32 respectively. The second intersections of circle AKB with AC and BC are 13 and 23 respectively. It is proved that these six points lie on a circle in which 12 13 is parallel to BC etc. and hence is a Tucker circle. Its centre P is shown to lie on the Brocard axis OK. When lines 12 13, 23 21 and 31 32 are extended a triangle UVW is formed whose centre Q is also shown to lie on the Brocard axis. When the sides of UVW are extended the second intersections of VW with circles CKA and AKB are points V1, W1 respectively. Similarly W2 and U2 lie on WU and U3 and V3 lie on UV. It is shown that these six points also lie on a circle, this time with centre O, the circumcentre of ABC. The construction can clearly be continued indefinitely creating an endless chain of circles. Areal co-ordinates, with ABC triangle of reference are used in following sections. 2. Circles BKC, CKA, AKB and their centres D, E, F Circles in areal co-ordinates have equations of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(2.1)
The point K has co-ordinates (a2, b2, c2). Circle BKC has v = w = 0 and u = – 3b2c2/(a2 + b2 + c2).
(2.2)
Circle CKA has w = u = 0 and v = – 3c2a2/(a2 + b2 + c2).
(2.3)
Circle AKB has u = v = 0 and w = – 3a2b2/(a2 + b2 + c2).
(2.4)
The centre of a circle written in the form of equation (2.1) has centre (x, y, z), where x = – (a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w)),
(2.5)
and y, z follow by cyclic change of a, b, c and u, v, w. 2
The centre D therefore has co-ordinates (x, y, z), where x = a2(b4 + c4 – a4 – 4b2c2), y = b2(a4 – b4 – 2c4 + 3b2c2 + 5c2a2), z = c2(a4 – 2b4 – c4 + 3b2c2 + 5a2b2).
(2.6) (2.7) (2.8)
The co-ordinates of the centres E and F may be written down from Equations (2.6) – (2.8) by cyclic change of x, y, z and a, b, c. 3. The six points 31, 21, 12, 32, 23, 13 lie on a circle centre P Circle BKC meets CA (y = 0) at the point 31 with co-ordinates (a2 + b2 – 2c2, 0, 3c2). Circle BKC meets AB (z = 0) at the point 21 with co-ordinates (c2 + a2 – 2b2, 3b2, 0). Circle CKA meets AB (z = 0) at the point 12 with co-ordinates (3a2, b2 + c2 – 2a2, 0). Circle CKA meets BC (x = 0) at the point 32 with co-ordinates (0, a2 + b2 – 2c2, 3c2). Circle AKB meets BC (x = 0) at the point 23 with co-ordinates (0, 3b2, c2 + a2 – 2b2). Circle AKB meets CA (y = 0) at the point 13 with co-ordinates (3a2, 0, b2 + c2 – 2a2). It may now be checked that the circle through these six points has an equation of the form (2.1) with u = 3b2c2(2a2 – b2 – c2)/(a2 + b2 + c2)2, (3.1) 2 2 2 2 2 2 2 2 2 v = 3c a (2b – c – a )/(a + b + c ) , (3.2) 2 2 2 2 2 2 2 2 2 w = 3a b (2c – a – b )/(a + b + c ) . (3.3) Using Equation (2.5) we find the x-co-ordinate of its centre P to be x = a2(a4 + 2(b4 + c4) – 2b2c2 – 3a2(b2 + c2)).
(3.4)
Its y- and z- co-ordinates follow by cyclic change of a, b, c. It may now be checked that P lies on the Brocard axis with equation b2c2(b2 – c2)x + c2a2(c2 – a2)y + a2b2(a2 – b2)z = 0.
(3.5)
4. The triangle UVW and the circle UVW centre Q The equation of the line 12 13 is (2a2 – b2 – c2)x + 3a2(y + z) = 0.
(4.1)
This is clearly parallel to BC and hence the circle of Section 3 is a Tucker circle and also it meets the line 21 23 at W with co-ordinates (3a2, 3b2, c2 – 2a2 – 2b2). U and V have co-ordinates that may be obtained from those of W by cyclic change of x, y, z and a, b, c.
3
Circle UVW has equation of the form (2.1) with u = 3b2c2(7a2 – 2(b2 + c2))/(a2 + b2 + c2)2,
(4.2)
with v, w obtained from Equation (4.2) by cyclic change of a, b, c. Using Equation (2.5) we find the x-co-ordinate of its centre Q to be x = a2(a4 + 5b4 + 5c4 – 2b2c2 – 6a2(b2 + c2)),
(4.3)
with the y- and z- co-ordinates obtained from (4.3) by cyclic change of a, b, c. It may now be checked that Q lies on the Brocard axis with Equation (3.5). It is also the case that AU, BV, CW obviously concur at K. 5. Points W1, V1, U2, W2, V3, U3 and the circle through these six points The line 12 13 meets circle AKB at the point W1 with co-ordinates (3a2, b2 + 4c2 – 2a2, – 3c2). The line 12 13 meets circle CKA at the point V1 with co-ordinates (3a2, – 3b2, c2 + 4b2 – 2a2). Similarly U2 has co-ordinates (– 3a2, 3b2, c2 +4a2 – 2b2), W2 has co-ordinates (a2 + 4c2 – 2b2, 3b2, – 3c2), V3 has co-ordinates (a2 + 4b2 – c2, – 3b2, 3c2) and U3 has co-ordinates ( – 3a2, b2 + 4a2 – 2c2, 3c2). It may now be shown that these six points lie on a circle of the form (2.1) with u = v = w = 9a2b2c2/(a2 + b2 + c2)2. As with all circles with u = v = w its centre is the circumcentre O.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
(5.1)
Article: CJB/153/2011
More properties of the Incentre Christopher Bradley Abstract: In a triangle ABC, with incentre I, the midpoints of AI, BI, CI are located. Perpendicular bisectors of AD, BE, CF are drawn. These meet the interiors of the sides of ABC at six points that are shown to lie on an ellipse. As the figure develops to include the excentres several other circles and conics are located.
Z
a2 b1 C' A 32 N
Q P O 2
B
a3
D W
31`
23
E
M
1 V I U
13
Y
3 F
21` 12
L
A' b3 c2 X
Fig. 1 1
C
B' c1
1. Introduction Given triangle ABC the incentre I is located and also the circumcentre O. Points D, E, F are the midpoints of AI, BI, CI respectively. EF meets side AB at 31 and side CA at 21. Points 12 and 32 lie on side FD where it intersects BC and AB respectively. Points 23 and 13 lie on side DE where it intersects CA and BC respectively. It is proved that thee six points lie on a conic. Lines 31 13 and 12 21 meet at a point A' on AI and points B' and C' are similarly defined. It is shown that the circle A'B'C' has its centre at a point P lying on IO. The points A', E, F, 13, 12 lie on a conic passing through points 2 on BI and 3 on CI. The points B', F, D, 23, 21 lie on a conic passing through 3 on CI and a point 1 on AI. The points C', D, E, 32, 31 lie on a conic passing through the previously defined points 1 on AI and 2 on BI. The lines through A, B, C perpendicular to AI, BI, CI form the triangle XYZ of excentres. It is shown that circle XYZ has centre Q also lying on IO. The line B'C' meets ZX at the point b1 and the point XY at the c1. Points c2, a2 lie on C'A' where it intersects XY and YZ respectively. Points a3 and b3 lie on A'B' where it intersects YZ and ZX respectively. It is shown these six points of intersection lie on a conic. 2. Triangle DEF Point D is the midpoint of AU and so has co-ordinates D(2a + b + c, b, c). Similarly E and F have co-ordinates E(a, 2b + c + a, c), F(a, b, 2c + a + b). It follows that EF has equation (a + 2(b + c))x – a(y + z) = 0.
(2.1)
(b + 2(c + a))y – b(z + x) = 0,
(2.2)
(c + 2(a + b))z – c(x + y) = 0.
(2.3)
Similarly FD, DE have equations
and
3. The six points 13, 12, 21, 23, 32, 31 and the conic on which they lie DE meets BC at 13 with co-ordinates (0, 2a + 2b + c, c). 2
FD meets BC at 12 with co-ordinates (0, b, 2c + 2a + b). EF meets CA at 21 with co-ordinates (a, 0, 2b + 2c + a). DE meets CA at 23 with co-ordinates (2a + 2b + c, 0, c). FD meets AB at 32 with co-ordinates (2c + 2a + b, b, 0). EF meets AB at 31 with co-ordinates (a, 2b + 2c + a, 0). It may now be shown that these six points lie on a conic with an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where u = bc(2b + 2c + a), v = ca(2c + 2a + b), w = ab(2a + 2b + c), f = – a(2a2 + b2 + c2 + 3bc + 3ca + 3ab), g = – b(2b2 + c2 + a2 + 3ca + 3ab + 3bc), h = – c(2c2 + a2 + b2 + 3ab + 3bc + 3ca).
(3.2)
The line AI meets EF at U with co-ordinates U(a(b + c), b(a + 2b + 2c), c(a + 2b + 2c)) and the conic with equation (3.2) at L with co-ordinates (x, y, z), where x = 2a2 + 3b2 + 3c2 + 6bc + 6ca + 6ab – (a + b + c)√(8a2 + 9b2 + 9c2 + 18bc + 20ca + 20ab), y = b(a + 2b + 2c), z = c(a + 2b + 2c). (3.3) Points V, M on BI and points W, N on CI have co-ordinates that may be obtained from those of U, L by cyclic change of x, y, z and a, b, c. 4. Points A', B', C' and circle A'B'C' with centre P on IO The line 23 32 has equation – bcx + c(2a + b + 2c)y + b(2a + 2b + c)z = 0.
(4.1)
The line 31 13 has equation c(a + 2b + 2c)x – cay + a(2a + 2b + c)z = 0.
(4.2)
These lines meet at the point C' with co-ordinates (x, y, z) = (a(2a + 2b + c), b(2a + 2b + c), – c(a + b + 2c)).
(4.3)
Points A', B' have co-ordinates that may be written down from those of C' by cyclic change of x, y, z and a, b, c. Note that it is immediate that A' lies on AI etc. Circle A'B'C' has an equation of the form a2yz + b2zx + c2xy + (fx + gy + hz)(x + y + z) = 0, where 3
(4.4)
f = bc(a + 2b + 2c)/(4(a + b + c)), g = ca(2a + b + 2c)/(4(a + b + c)), h = ab(2a + 2b + c)/(4(a + b + c)).
(4.5)
The x- co-ordinate of the centre of a circle in the form (4.4) is x = – (a4 – a2(b2 + c2 + 2f – g – h) + (b2 – c2)(g – h)).
(4.6)
It follows that P, the centre of circle A'B'C', has x co-ordinate x = – (a/2)(2a3 + a2(b + c) – 2a(b2 + bc + c2) – (b2 + c2)(b – c)).
(4.7)
It may now be checked that P lies on IO with equation bc(b – c)(b + c – a)x + ca(c – a)(c + a – b)y + ab(a – b)(a + b – c)z = 0.
(4.8)
5. The conic A' 13 12 E F and the point 2 (and others by cyclic change) It may now be checked that the five points A', 13, 12, E, F lie on a conic with equation of the form (3.1), where u = bc(a + 2b + 2c)(3a2 + 2b2 + 2c2 + 4bc + 6ca + 6ab), v = a2c(a + 2b + 2c)((2a + b + 2c), w = a2b(a + 2b + 2c)(2a + 2b + c), f = – a2(a + 2b + 2c)(2a2 + b2 + c2 + 3bc + 3ca + 3ab), g = ab(a3 + 3a2b + a(b + c)(2b – 3c) – c(b + c)2), h = ac(a3 + 3a2c + a(b + c)(2c – 3b) – b(b + c)2). (5.1) The conics B' 21 23 F D and C' 32 31 D E now have equations that may be written down by cyclic change of u, v, w, and a, b, c, and f, g, h. The conic A' 13 12 E F meets the line BI at the point 2 with co-ordinates (x, y, z), where x = a(a + 2b + 2c)(2a + b + 2c), y = b(5a2 + 2b2 + 4c2 + 6bc + 9ca + 8ab), (5.2) z = c(a + 2b + 2c)(2a + b + 2c). Co-ordinates of points 3, 1 may be obtained from (5.2) by cyclic change of x, y, z and a, b, c. 6. Points X, Y, Z and circle XYZ, centre Q on IO The points X, Y, Z have co-ordinates X(– a, b, c), Y(a, – b, c), Z(a, b, – c) and it is now easy to find the equation of circle XYZ. It has the form of Equation (4.4) with f = bc, g = ca, h = ab. The x- co-ordinate of its centre Q (using Equation (4.6)) is x = – a(a3 + a2(b + c) – a(b + c)2 – (b2 – c2)(b – c)). (6.1)
4
the y- and z- co-ordinates follow from Equation (6.1) by cyclic change of a, b, c. It may now be checked that Q lies on IO, with Equation (4.8). 7. Points a2, a3, b3, b1, c1, c2 and the conic on which they lie The line YZ has equation cy + bz = 0. The line C'A' has equation c(a + 2b + 2c)x – cay + a(2a + 2b + c)z = 0.
(7.1)
These two lines meet at the point a2, which therefore has co-ordinates (x, y, z), where x = a(2a + 3b + c), y = b(a + 2b + 2c), z = – c(a + 2b + 2c).
(7.2)
The line A'B' has equation b(a + 2b + 2c)x + a(2a + b + 2c)y – abz = 0.
(7.3)
The intersection of YZ and A'B' is the point a3, which therefore ha co-ordinates (x, y, z), where x = a(2a + b + 3c), y = – b(a + 2b + 2c), z = c(a + 2b + 2c). (7.4) The co-ordinates of b3 and c1 follow from those of a2 by cyclic change of x, y, z and a, b, c and the co-ordinates of b1 and c2 follow from those of a3 by cyclic change of x, y, z and a, b, c. It is now possible to work out the equation of the conic through a2, a3, b3, b1, c1, c2 and it has the form of equation (3.1) with u = b2c2(a + 2b + 2c), (7.5) with v, w following from u by cyclic change of a, b, c and f = a2bc(6a2 + 4b2 + 4c2 + 9bc + 10ca + 10ab), with g, h following from f by cyclic change of a, b, c.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
(7.6)
6
Article: CJB/2011/154
On the Incircle and Excircles of a Cevian Triangle Christopher Bradley Abstract: Given triangle ABC, if D, E, F are the feet of the Cevians AD, BE, CF and L, M, N are the points of contact of the incircle of triangle DEF with the sides of triangle DEF then AL, BM, CN are concurrent, as are AL', BM', CN', where L', M', N' are the points of contact of the excircles with EF, FD, DE respectively.
A
W' F
V' L'
W
E
R' Q
M'
P
V N'
R
M
B
L
N
U' U
D
C
Fig. 1 The Cevian triangle DEF and its incircle and ex-circles
1
1. Introduction Given any triangle ABC and a point P not on its sides, let AP, BP, CP meet BC, CA, AB respectively at D, E, F. Then we define triangle DEF to be the Cevian triangle of P. Its incircle, centre Q, touches D, E, F respectively at L, M, N. We prove AL, BM, CN are concurrent at a point R. If the three ex-circles touch EF, FD, DE respectively at L', M', N' then we prove AL', BM', CN' are also concurrent at a point R'. We use areal co-ordinates with DEF as triangle of reference. Four conics are also identified. 2. Triangle ABC and its sides, the point P and the Cevians APD, BPE, CPF We take DEF to be the triangle of reference as this leads to easier algebra than if ABC is chosen. We start with three parameters r, s, t and then eliminate r to take account of the fact that AD, BE, CF are concurrent. We take BC, CA, AB to have equations y + tz = 0, z + sx = 0, x + ry = 0, where r, s, t are positive parameters. (This is equivalent to choosing P to be an internal point of triangle DEF, which is a mild restriction.) Points A, B, C therefore have co-ordinates A(– r, 1, rs), B(rt, – t, 1), C(1, st, – s). In consequence AD, BE, CF have equations z = rsy, x = rtz, y = stx respectively. Since these lines are concurrent we have rst = 1 and so we set r = 1/st and modify the above co-ordinates and equations accordingly, which are now as follows: BC: y + tz = 0, CA: z + sx = 0, AB: y + stx = 0. (2.1) A (– 1, st, s), B(1, – st, s), C(1, st, – s), P(1, st, s). (2.2) AD: y = tz, BE: z = sx, CF y = stx. (2.3) 3. The incircle, points L, M, N and the concurrency of AL, BM, CN at R With a, b, c the lengths of the sides of EF, FD, DE respectively, the incircle of triangle DEF has equation (b + c – a)2x2 + (c + a – b)2y2 + (a + b – c)2z2 – 2(a + b – c)(c + a – b)yz – 2(b + c – a)(a + b – c)zx – 2(c + a – b)(b + c – a)xy = 0. (3.1) This touches EF at L(0, a + b – c, c + a – b). Similarly M has co-ordinates M(a + b – c, 0, b + c – a), and N(c + a – b, b + c – a, 0). The equation of AL is 2
(3.2)
s((b – c)(t + 1) – a(t – 1))x – (c + a – b)y + (a + b – c)z = 0
(3.3)
Likewise the equation of BM is st(b + c – a)x – (a(s + 1) + b(s – 1) – c(s + 1))y – st(a + b – c)z = 0.
(3.4)
And the equation of CN is s(b + c – a)x – s(c + a – b)y – (a(st + 1) – b(st + 1) + c(st – 1))z = 0.
(3.5)
It may now be checked that AL, BM, CN are concurrent at a point R with co-ordinates (x, y, z), where x = a(st + s + 1) – b(st – s + 1) + c(st – s – 1), y = – st(a(st – s + 1) – b(st + s + 1) + c(st + s – 1)), (3.6) z = s(a(st – s – 1) + b(1 – st – s) + c(st + s + 1)). 4. The points L', M', N' and the concurrency of AL', BM', CN' at R' Given the co-ordinates of L, M, N those of L', M', N' follow immediately and are L'(0, c + a – b, a + b – c), M'(b + c – a, 0, a + b – c), N'(b + c – a, c + a – b, 0). The equation of AL' is – s((a(t – 1) + (b – c)(t + 1))x – (a + b – c)y + (c + a – b)z = 0.
(4.1)
Similarly the equation of BM' is st(a + b – c)x + (a(s + 1) + b(1 – s) – c(s + 1))y – st(b + c – a)z = 0.
(4.2)
And the equation of CN' is – s(c + a – b)x + s(b + c – a)y – (a(st + 1) – b(st + 1) + c(1 – st))z = 0.
(4.3)
It may now be checked that AL', BM', CN' are concurrent at a point R' with co-ordinates (x, y, z), where x = a2(s(t + 1) + 1) – 2as(b + ct) – b2(s(t – 1) + 1) + 2bc + c2(s(t – 1) – 1), y = – st(a2(s(t – 1) + 1) + 2as(b – ct) + (c – b)(b(s(t + 1) + 1) + c(s(t + 1) – 1))), z = s(a2(s(t – 1) – 1) + 2as(b – ct) + b2(1 – s(t + 1)) – 2bc + c2(s(t + 1) + 1)). (4.4) 5. The four conics Let AR, BR, CR meet BC, CA, AB respectively at U, V, W, and let AR', BR', CR' meet BC, CA, AB respectively at U', V', W'. Four conics may now be identified. There existence is well-known and follow from properties of in-circles, ex-circles, and sets of points that are the feet of pairs of 3
Cevians. Their equations are not important so we merely record their existence and they can be seen in Fig. 1. They are (i) L, M, N, L', M', N'; (ii) D, E, F, U, V, W; (iii) D, E, F, U', V', W'; (iv) U, V, W, U', V', W'.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/155
The Circumconic of a pair of Cevian Triangles Christopher Bradley Abstract: In triangle ABC two Cevian triangles are drawn and their points of intersection and the vertices of ABC lead to six sets of collinear points. Several conics are identified.
22
A
33
12 N W 21
R
13 P
23
U
31
M
Q
B
V
11
X'
L
C
To S 32
Fig. 1 Cevian triangles LMN and UVW and six sets of collinear points 1
1. Introduction Given triangle ABC and two points P and Q not on the sides then the triangles LMN and UVW formed by the feet of the Cevians of P and Q respectively are drawn. Sides MN and VW intersect at the point 11, sides NL and WU intersect at point 22 and sides LM and UV meet at the point 33. Point 12 is the intersection of MN and WU, and point 21 is the intersection of NL and VW. Point 23 is the intersection of NL and UV, and point 32 is the intersection of LM and WU. Finally point 31 is the intersection of LM and VW and 13 is the intersection of MN and UV. It is proved the following sets of three points are collinear: (a) A 22 33; (b) B 33 11; (c) C 11 22; (d) A 23 32; (e) B 31 13; (f) C 12 21. It is also shown that the lines (d), (e), (f) are concurrent at a point R and also that A 11, B 22, C 33 are concurrent at S, so that triangles ABC and 11 22 33 are in perspective with vertex S. It is also shown that a conic passes through L, U, 12, 21, 31, 13 and that there are two similar conics by cyclic change of symbols. Areal co-ordinates are used with ABC the triangle of reference. 2. Points L, M, N, U, V, W and the sides of triangles LMN and UVW We take P to have co-ordinates P(l, m, n). Then L, M, N have co-ordinates L(0, m, n), M(l, 0, n) and N(l, m, 0). The sides of triangle LMN have the following equations: MN: – mnx + nly + lmz = 0. (2.1) NL: mnx – nly + lmz = 0. (2.2) LM: mnx + nly – lmz = 0. (2.3) We take Q to have co-ordinates Q(u, v, w). The U, V, W have co-ordinates U(0, v, w), V(u, 0, w) and W(u, v, 0). The sides of triangle UVW have the following equations: VW: – vwx + wuy + uvz = 0. (2.4) WU: vwx – wuy + uvz = 0. (2.5) UV: vwx + wuy – uvz = 0. (2.6) It is well-known that the feet of a pair of Cevians are co-conic and its equation may be checked to be of the form px2 + qy2 + rz2 + 2fyz + 2gzx + 2hxy = 0, (2.7) where p = 2mnvw, q = 2nlwu, r = 2lmuv, f = – lu(mw + nv), g = – mv(nu + lw), h = – nw(lv + mu). (2.8) 3. The points 11, 22, 33 and lines A 22 33, B 33 11, C 11 22 and the point S The point 11 = MN^VW and has co-ordinates 11(lu(nv – mw), mv(nu – lw), nw(lv – mu)). 2
The point 22 = NL^WU and has co-ordinates 22(lu(mw – nv), mv(lw – nu), nw(lv – mu)). The point 33 = LM^UV and has co-ordinates 33(lu(nv – mw), mv(lw – nu), nw(lv – mu)). It may now be shown that line 22 33 has equation nw(lv – mu)y + mv(nu – lw)z = 0,
(3.1)
a line which obviously passes through A. Similarly 33 11 passes through B and 11 22 passes through C. The line A 11 has equation nw(lv – mu)y = mv(nu – lw)z , (3.2) The equations of the lines B 22 and C 33 may be obtained from (3.2) by cyclic change of x, y, z and l, m, n and u, v, w. It may now be shown that these three lines meet at a point S with coordinates (x, y, z), which are, unsurprisingly, x = lu(mw – nv), y = mv(nu – lw), z = nw(lv – mu). (3.3) 4. The points 23, 32 and the line 23 32 passing through A and R The point 23 = NL^UV and has co-ordinates 23(lu(nv – mw), mv(lw + nu), nw(mu + lv)). The point 32 = LM^WU and has co-ordinates 32(lu(mw – nv), mv(lw + nu), nw(mu + lv)). It may now be shown that line 23 32 has equation nw(mu + lv)y = mv(lw + nu)z.
(4.1)
It is evident that 23 32 passes through A and also the point R with co-ordinates (x, y, z), where x = lu(mw + nv), y = mv(nu + lw), z = nw(lv + mu). (4.2) The cyclic nature of these co-ordinates shows that R also lies on 31 13 and 12 21, which in turn pass through B and C respectively. 5. Three conics, one passing through W, N, 32, 23, 13, 31 Point 31 = LM^VW and has co-ordinates 31(lu(nv + mw), mv(lw – nu), nw(mu + lv)). Point 13 = MN^UV and has co-ordinates 13(lu(mw + nv), mv(nu – lw), nw(lv + mu)). From the co-ordinates of points W, N, 32, 23, 13, 31 we may conclude that they lie on a conic of the form (2.7) with p = – 2mn2vw2 (lv + mu), (5.1) 2 2 q = – 2ln uw (lv + mu), (5.2) r = 2lmuv(lw(2mw + nv) + nu(mw + 2nv)), (5.3) f = – lnuw(lv + mu)(mw + nv), (5.4) 3
g = – mnvw(lv + mu)(nu + lw), h = n2w2(lv + mu)2.
(5.5) (5.6)
Similarly conics also pass through (i) U, L, 13, 31. 21, 12, (ii) V, M, 21, 12, 32, 23. See Fig. 1
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Article: CJB/2011/156
Circumcentre Conics Christopher Bradley Abstract: Given triangle ABC, let triangle UVW be a reduction of ABC by an enlargement factor t (0 < t < 1) about the circumcentre O. The sides of triangle UVW cut the sides of ABC in six points that lie on a conic. It is shown that the centres of all such conics lie on the Brocard axis.
A 23 32
U
N
M
T E O
31
K
S
G
V
W
21
R B
13
L
12
C
Fig. 1 A circumcentre conic, centre E lying on the Brocard axis OK
1
1. Introduction In Fig. 1 triangle UVW is an enlargement/reduction of triangle ABC with centre of enlargement the circumcentre O. It follows that the sides of UVW are parallel to those of ABC and that the circumcentre of UVW is also the point O. Side VW meets AB at the point 31 and the side CA at the point 21. Side WU meets BC at the point 12 and the side AB at point 32. Side UV meets side CA at the point 23 and the side BC at the point 13. It is proved that these six points lie on a conic and that whatever the enlargement factor the centre E always lies on the Brocard axis OK, where K is the symmedian point of triangle ABC. Further properties are also established: (i) B, V, W, C are concyclic, (ii) C, W, U, A are concyclic, (iii) A, U, V, B are concyclic, (iv) If B 21 meets C 31 at R, then AR is the median through A, (v) If C 32 meets A 12 at S, then BS is the median through B, and (vi) If A 13 meets B 23 at T, then CT is the median through C. 2. Points U, V, W and the sides of triangle UVW We suppose that the x-co-ordinate of U is (1 – k) times that of A plus k times that of O. Then x= (1/s)(a4 + (k – 2)a2(b2 + c2) + (1 – k)(b2 – c2)2), (2.1) where s = (a4 + b4 + c4 – 2b2c2 – 2c2a2 – 2a2b2). Similarly the y- and z- co-ordinates are y = – (1/s)(b2k(c2 + a2 – b2)), z = – (1/s)(c2k(a2 + b2 – c2)).
(2.2) (2.3)
The co-ordinates of V and W follow from those of U by cyclic change of x, y, z and a, b, c. It is now possible to determine the equations of the sides of triangle UVW. That of VW is (a4(k – 1) + a2(2 – k)(b2 + c2) – (b2 – c2)2)x + a2k(a2 – b2 – c2)(y + z) = 0. (2.4) The equations of WU and UV may be obtained from (2.4) by cyclic change of x, y, z and a, b, c. 3. The points 31, 21, 12, 32, 23, 13 and the conic through these six points Point 31 = VW^AB has co-ordinates (x, y, z), where x = a2k(b2 + c2 – a2), y = a4(k – 1) + a2(b2 + c2)(2 – k) – (b2 – c2)2, z = 0.
(3.1) (3.2) (3.3)
Point 21 = VW^CA has co-ordinates (x, y, z), where x = a2k(b2 + c2 – a2), y = 0, z = a4(k – 1) + a2(b2 + c2)(2 – k) – (b2 – c2)2.
(3.4) (3.5) (3.6)
2
The co-ordinates of 12, 23 may be obtained from those of 31 by cyclic change of x, y. z and a, b, c. The co-ordinates of 32, 13 may be obtained from those of 21 by cyclic change of x, y, z and a, b, c. It is now possible to determine the equation of the conic passing through these six points. (The three other points of intersection of the two triangles lie on the line at infinity and so the six points detailed above must lie on a conic.) Its equation is of the form px2 + qy2 + rz2 + 2fyz + 2gzx + 2hxy = 0,
(3.7)
where p = – 2b2c2k(a2 + b2 – c2)(c2 + b2 – a2)(a4(k – 1) + a2(2 – k)(b2 + c2) – (b2 – c2)2),
(3.8)
and q, r follow from (3.8) by cyclic change of a, b, c and f = a2(b2 + c2 – a2)(a8 + a6(b2 + c2)(k – 4) – a4(3b4(k – 2) – 2b2c2(k2 – k + 2) + 3c4(k – 2)) + a2(b2 + c2)(b2 – c2)2(3k – 4) – (b2 – c2)2(b4(k – 1) + 2b2c2(k2 – k + 1) + c4(k – 1)), (3.9) and g, h follow from (3.9) by cyclic change of a, b, c. The centre E of a conic in the form of (3.7) has centre given by x = qr – gq – hr – f2 + fg + hf, y = rp – hr – fp – g2 + gh + fg, z = pq – fp – gq – h2 + hf + gh.
(3.10)
Unfortunately the centre E in this case has co-ordinates of degree 24 in a, b, c, though oddly enough linear in k (DERIVE). It is possible then to use algebraic software to determine the locus of E as k varies and this turns out to be (b2 – c2)b2c2x + (c2 – a2)c2a2y + (a2 – b2)a2b2z = 0, (3.11) which is the equation of the Brocard axis. I suppose this would have been an intelligent guess, since if you start with K rather than O, then the ‘Symmedian Conics’ are circles with centres also lying on the Brocard axis. 4. Circles such as BVWC The equation of the circle passing through B, C, V, W is of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0, with v = w = 0 and u = b2c2(2 – k)/(a2 – b2 – c2). 3
(4.1)
Circles CWUA and AUVB follow from equation (4.1) by cyclic change of x, y, z and a, b, c and u, v, w. 5. The point R = B 21^C 31 lies on the median through A The equation of B 21 is a2k(b2 + c2 – a2)z = (a4(k – 1) + a2(2 – k)(b2 + c2) – (b2 – c2)2) x.
(5.1)
The equation of C 31 is a2k(b2 + c2 – a2)y = (a4(k – 1) + a2(2 – k)(b2 + c2) – (b2 – c2)2) x.
(5.2)
These lines obviously meet on the line y = z, which is the median through A. Similarly S lies on the median through B and T lies on the median through C.
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Article: CJB/2011/157
The General Inellipse Christopher Bradley Abstract: The general inellipse in a triangle touches its sides at the feet of Cevians. An analysis is given that relates the Cevian point to the centre of the inellipse and to two related circumconics.
A
C'
B'
T
S
N M
O P
L
B
C
A' R
Fig. 1 An inellipse, its centre and two related circumconics
1
1. Introduction In Fig. 1 P is an internal point of a triangle ABC and the feet of the Cevians AP, BP, CP are L, M, N respectively. The inellipse touches the sides BC, CA, AB at the points L, M, N. The point O is the centre of the inellipse. The two circumconics pass through A, B, C and have centres O and P. An analysis is given of these results in terms of the co-ordinates of P, using areal coordinates with ABC as triangle of reference. (When P is external one obtains an ex-conic, which may be a hyperbola.) 2. An inellipse and its centre The equation of an inellipse is well-known to be of the form u2x2 + v2y2 + w2z2 – 2vwyz – 2wuzx – 2uvxy = 0,
(2.1)
where u, v, w are positive constants. It touches BC at L (0, 1/v, 1/w), CA at M (1/u, 0, 1/w) and AB at N (1/u, 1/v, 0). Clearly L, M, N are the feet of Cevians AP, BP, CP respectively, where P has co-ordinates (1/u, 1/v, 1/w). The centre O of the conic with equation (2.1) has co-ordinates (v + w, w + u, u + v). For the incircle u = (1/2)(b + c – a), v = (1/2)(c + a – b), w = (1/2)(a + b – c), with incentre I (a, b, c) and P Gergonne’s point Ge (1/(b + c – a), 1/(c + a – b), 1/(a + b – c). 3. The circumconic centre O Let A', B', C' be the images of A, B, C after rotation by 180o about O. Then the co-ordinates of A', B', C' are respectively A'(– u, w + u, u + v), B'(v + w, – v, u + v), C'(v + w, w + u, – w). It may now be shown that the circumconic ABCA'B'C', centre O, has an equation of the form fyz + gzx + hxy = 0, (3.1) where f = u(v + w), g = v(w + u), h = w(u + v). For the incircle f = a(b + c – a), g = b(c + a – b), h = c(a + b – c). 4. The circumconic centre P Let R, S, T be the images of A, B, C after rotation by 180o about P. Then the co-ordinates of R, S, T are respectively R(vw – wu – uv, 2wu, 2uv), S(2vw, wu – uv – vw, 2uv), T(2vw, 2wu, uv – vw – wu). It may now be shown that the circumconic ABCRST, centre P, has an equation of the form (3.1) with 2
f = vw(u(v + w) – vw), g = wu(v(w + u) – wu), h = uv(w(u + v) – uv). For the incircle f, g, h are not memorable.
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3
(4.1)
Article: CJB/2011/158
Six Point Circles and their Associated Conics Christopher Bradley Abstract: If P is a point internal to a triangle ABC and AP, BP, CP meet the circumcircle again at A', B', C' respectively and U, V, W are the midpoints of AP, BP, CP and D, E, F are the midpoints of PA', PB', PC' respectively then U, V, W, D, E, F are obviously concyclic. But the configuration has other properties which are pointed out.
A
A'' C'
Co
Bo
U 3a O
F
2a 13
3b
Q
B'
P
S R
23
12 E 32
B'' 21
V B
31
W 1c
1b
2c
C
D
C'' A' Ao
Fig. 1 The six-point circle and associated conics 1
1. Introduction In Fig. 1 ABC is a triangle and P an internal point. AP, BP, CP meet the circumcircle Σ of ABC again at points A', B', C'. Points U, V, W are the midpoints of AP, BP, CP and D, E, F are the midpoints of PA', PB', PC' respectively. It is proved that U, V, W, D, E, F are concyclic. The six intersections inside Σ of the triangles ABC and A'B'C' (labelled 1b, 1c, 2c, 2a, 3a, 3b are shown to be co-conic). Triangles UVW and DEF are homothetic with triangles ABC and A'B'C'(being half their size) and so their six intersections (labelled 21, 31, 322, 12, 13, 23) are also co-conic. The centres of the circles are collinear with P, as are the centres of the ellipses. The lines through U, V, W parallel to the sides BC, CA, AB form a triangle A oBoCo and U is the midpoint of BoCo etc. Similarly the lines through D, E, F parallel to the sides B'C', C'A', A'B' form a triangle A''B''C'' and D is the midpoint of B''C''. Most of the results above follow from the homothety whereby triangle ABC is mapped onto triangle DEF by a reduction through P by a scale factor of 2. However, besides using this homothety we choose to give some analytic results (which show how complicated analytic geometry can sometimes be compared with pure geometry). 2. The circle UVWDEF The three chords AA', BB', CC' meet at P so AP.PA' = BP.PB' = CP.PC' = k, say. It follows that UP.PD = VP.PE = WP.PF each being ¼k. Since the centre of reduction is P it also follows that PQ= QO, where O is the circumcentre of ABC and Q is the circumcentre of circle UVWDEF. We now present the key results of a calculation using areal co-ordinates with ABC as triangle of reference. The unnormalised co-ordinates of A' are (– a2gh, g(b2h + c2g), h(b2h + c2g)). Those of B' and C' follow by cyclic change of x, y, z and f, g, h and a, b, c. The unnormalised co-ordinates of U are (1 + f, g, h) and those of V and W follow by cyclic change of x, y, z and f, g, h. The unnormalised co-ordinates of D are (a2gh(f + 1) – f(g + h)(b2h + c2g), g(a2gh – (g + h + 1)(b2h + c2g)), h(a2gh – (g + h + 1)(b2h + c2g)). (2.1) It may now be checked that the equation of the circle UVWDEF is of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0,
(2.2)
where u = ((a2gh – (g + h + 1)(b2h + c2g)))/(f + g + h + 1)2, v = (f(b2h – c2(h + f + 1)) – a2h(h + f + 1))/(f + g + h + 1)2, 2
(2.3) (2.4)
w = – (a2g (f + g + 1) + f (b2(f + g + 1) – c2g))/(f + g + h + 1)2 The centre of a circle with equation of the form (2.2) has x-co-ordinate x = – (a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w)).
(2.5)
(2.6)
The y- and z- co-ordinates follow by cyclic change of a, b, c and u, v, w. The co-ordinates of Q are therefore (x, y, z), where x = a4 (f + 1) – a2(b2 + c2)(2f + 1) + f(b2 – c2)2,
(2.7)
and y, z follow by cyclic change of a, b, c and f, g, h. It may now be checked that Q is the midpoint of OP, where O is the circumcentre of Σ. 3. The Associated Conics Consider triangles ABC and A'B'C'. There are 9 points of intersection. One of two things can happen. Either the 9 points lie on a general cubic curve degree 3) or 3 of them are collinear (degree 1) and the remaining 6 lie on a conic (degree 2). We show that the three intersections BC^B'C', CA^C'A', AB^A'B' lie on a line. It follows that the six intersections 1b, 1c, 2c, 2a, 3a, 3b lie on a conic. The three points of intersection are: BC^B'C': x = 0, y = – b2f – a2g, z = c2f + a2h. CA^C'A': x = a2g + b2f, y = 0, z = – c2g – b2h. AB^A'B': x = – a2h – c2f, y = b2h + c2g, z = 0.
(3.1) (3.2) (3.3)
It may now be checked that these points lie on the line with equation (b2h + c2g)x + (c2f + a2h)y + (a2g + b2f)z = 0.
(3.4)
And hence the other six points of intersection lie on a conic. The homothety now ensures that the six points 21, 31, 32, 12, 13, 23 also lie on a conic. Also that if the centre of the first conic is S and the centre of the second conic is R, then PR = RS. 4. Another property of U, V, W The equation of the line through U parallel to BC is (g + h)x – (f + 1)(y + z) = 0.
3
(4.1)
The equation of the line through V parallel to CA is (h + f)y – (g + 1)(z + x) = 0.
(4.2)
The equation of the line through W parallel to AB is (f + g)z – (h + 1)(x + y) = 0.
(4.3)
The point of intersection Co therefore has co-ordinates (f + 1, g + 1, h – 1) and the point of intersection Bo has co-ordinates (f + 1, g – 1, h + 1). The midpoint of BoCo is evidently (f + 1, g, h), that is the point U. Similarly V is the midpoint of of CoAo and W is the midpoint of AoBo. It follows that circle UVW is the nine-point circle of triangle AoBoCo. It therefore passes through 6 more well-defined points, in addition to U, V, W, D, E, F.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article: CJB/2011/159
A Twelve Point Configuration and Carnot’s Theorem Christopher Bradley .
4' A
5 1'
4
U
F
15 45 P
24
5'
52 E
14
6
X
46 61
35
Y R
Q
3 2'
23
V 36
W
13
26 D
B
1
2
C
6'
3'
Fig. 1 A conic intersecting all three sides of a triangle gives rise to a wonderful configuration 1
Abstract: A twelve point configuration occurs when six of the points lie on a conic and the remaining six points are the vertices of two triangles in perspective. In this article we consider the configuration in which a conic cuts the three sides of a triangle in real points. A number of theorems create a variety of conditions on the points of intersection equivalent to Carnot’s theorem. 1. Introduction The construction of Fig.1 proceeds as follows: Given a conic Σ cutting BC at points 1, 2, CA at 3, 4 and AB at 5, 6 then determine the points of intersection such as 13, the intersection of A1 and B3. Altogether there are twelve such points 13, 26, 35, 24, 15, 46, 61, 36, 23, 52, 45, 14. We establish the result that the first six lie on a conic Γ and the lines 61 52, 36 45 and 23 14 are concurrent at a point we label X. But that is not the end of the matter; four other points of concurrence exist. There is P = 36 45^15 26^24 13, Q = 61 52^35 46^ 13 24, R = 14 23^15 26^35 46 and finally Y = AP^BQ^CR. We establish these results and their converses, showing them to be all equivalent to Carnot’s theorem that (B1/C1)(B2/C2)(C3/A3)(C4/A4)(A5/B5)(A6/B6) = 1 (1.1) It appears that the configuration can be developed further (to an infinite extent). For example the tangent to Γ at 15 meets AP at U and CY at F, with V, W and D, E similarly defined. The F also lies on the tangent at 46 and U lies on the tangent at 24 and so on. Then the tangents at 15 and 13 meet at a point 1', the tangents at 46 and 26 meet at 6' and so on and then 1', 2', 3', 4', 5', 6' are coconic. We do not establish these latter results analytically, as the article is long enough as it is, but the results are confirmed by CABRI II plus. See Fig. 1. 2. Points 1, 2, 3, 4, 5, 6 and the conic Σ We take points 1, 3, 5 to have co-ordinates 1(0, l, 1 –l), 3(1 – m, 0, m), 5(n, 1 – n, 0) and points 2, 4, 6 to have co-ordinates 2(0, 1 – p, p), 4(q, 0, 1 – q), 6(1 – r, r, 0). The conic Σ has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (2.1) and putting in the conditions that points 1, 2, 3, 4, 5 lie on Σ we obtain u = 2lmn(n – 1)(p – 1)(q – 1), (2.2) v = 2npq(l – 1)(m – 1)(n – 1), (2.3) w = 2nlq(m – 1)(n – 1)(p – 1), (2.4) f = nq(m – 1)(n – 1)(2lp – l – p + 1), (2.5) g = nl(n – 1)(p – 1)(2mq – m – q + 1), (2.6) 2 2 2 h = l(m(n (2pq – p – q + 1) – 2npq + pq) – pq(n – 1) ) – pq(m – 1)(n – 1) . (2.7) 2
After some heavy algebra we find the condition that point 6 lies on Σ is that r/(1 – r) = {lmn(1 – q)(1 – p)}/{pq(1 – l)(1 – m)(1 – n)}.
(2.8)
In fact what we have now done is to verify Carnot’s theorem (1.1). 3. The six points of intersection 13, 26, 35, 24, 15, 46 and the conic on which they lie The lines A1, B3, C5 have equations (1 – l)y = lz, (1 – m)z = mx, (1 – n)x = ny respectively. The lines A2, B4, C6 have equations py = (1 – p)z, qz = (1 – q)x, rx = (1 – r)y respectively. From equation (2.8) the equation of C6 is now taken as lmn(1 – p)(1 – q)x = pq(1 – l)(1 – m)(1 – n)y. (3.1) We now give the co-ordinates of six of the points of intersection of these lines: 13 = A1^B3: x = (1 – l)(1 – m), y = lm, z = m(1 – l). 35 = B3^C5: x = n(1 – m), y = (1 – m)(1 – n), z = mn. 15 = A1^C5: x = nl, y = l(1 – n), z = (1 – n)(1 – l). 26 = A2^C6: x = (1 – r)(1 – p), y = (1 – p)r, z = rp. 24 = A2^B4: x = pq, y = (1 – p)(1 – q), z = p(1 – q). 46 = B4^C6: x = q(1 – r), y = qr, z = (1 – q)(1 – r).
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
It may now be shown that these six points lie on a conic if, and only if, Equation (2.8) holds and then the conic has the form (2.1) with u = 2lmn(1 – n)(1 – q)2(l + p – 1)(m(n(2p – 1) – p) + p(1 – n)), (3.8) v = 2np2(1 – l)(1 – m)(1 – n)(m + q – 1)(l(n(2q – 1) – q) + q(1 – n)), (3.9) 2 2 2 w = 2l(m – 1)(l(m(n (p + q – 1) – 2npq + pq) – pq(n – 1) ) – pq(m – 1)(n – 1) ) all times (n(p(2q – 1) – q + 1) – pq), (3.10) 2 f = p(m – 1)(l (m + n – 1)(2qn – n – q) + l(n – 1)(m(n(2q – 1) – 2q) + n(1 – 3q) + 2q) – q(m – 1)(n – 1)2(pq – n(p(2q – 1) – q + 1)), (3.11) g = l(1 – q)(l(m + n – 1) + (1 – m)(1 – n))(m(n(2p – 1) – p) – p(n – 1)) all times (n(p(2q – 1) – q + 1) – pq), (3.12) 2 h = p(1 – q)(l (m + n – 1)(2nq – n – q) + l(n – 1)(m(n(2q – 1) – 2q) + n(1 – 3q) + 2q) – q(m – 1)(n – 1)2(m(n(2p – 1) – p – p(n – 1)). (3.13) 4. The six points of intersection 61, 36, 23, 52, 45, 14 and the point X We now give the co-ordinates of the other six points of intersection: 14 = A1^B4: x = q(1 – l), y = l(1 – q), z = (1 – l)(1 – q), 61 = C6^A1: x = l(1 – r), y = lr, z = r(1 – l), 36 = B3^C6: x = (1 – m)(1 – r), y = r(1 – m), z = m(1 – r), 23 = A2^B3: x = p(1 – m), y = m(1 – p), z = mp, 3
(4.1) (4.2) (4.3) (4.4)
52 = C5^B2: x = n(1 – p), y = (1 – n)(1 – p), z = p(1 – n), 45 = B4^C5: x = nq, y = q(1 – n), z = n(1 – q).
(4.5) (4.6)
The equation of 23 14 is m(l + p – 1)(q – 1)x + p(1 – l)(m + q – 1)y – (l(m(p – q) + p(q – 1)) + mq(1 – p)) = 0. (4.7) The equation of 36 45 is (m(n(q – r) + q(r – 1)) + nr(1 – q))x + n(1 – r)(m + q – 1)y + q(m – 1)(n + r – 1)z = 0. (4.8) The equation of 61 52 is r(1 – n)(l + p – 1)x +(l(n(p – r) + p(r – 1)) + nr(1 – p))y + l(n + r – 1)(p – 1)z = 0. (4.9) It may now be shown that these three lines are concurrent at a point X if, and only if, equation (2.8) holds. The co-ordinates of X are (x, y, z), where (with r given by (2.8)), x = – l(m(n(p(q + r – 1) – qr – r + 1) + pq(r – 1)) + n(p(q(r – 2) – r + 1) + q + r – 1) + pq(1 – r)) + mnqr(p – 1) + nqr(1 – p), (4.10) y = l(r(n(p(q – 1) + 1) – q) – m(n(p(q – r) + q(r – 1) + r) + q(p(r – 1) – 2r + 1))) + qr(1 – p)(1 – m)(1 – n), (4.11) z = l(m(n(p(q – r) – r(q – 1)) + pq(r – 1)) + nr(1 – q)(p – 1)) + nr(1 – p)(1 – q)(1 – m). (4.12) 5. Lines 24 13, 36 45, 15 26, are concurrent at P and Points Q, R and Y The equation of the line 24 13 is m(l + p – 1)(q – 1)x + p(l – 1)(m + q – 1)y + (l(m(p + q – 1) + (p – 1)(q – 1)) + (m – 1)(p – 1)(q – 1)) = 0. (5.1) The equation of the line 36 45 is (m(n(q – r) + q(r – 1)) + nr(1 – q))x + n(1 – r)(m + q – 1)y + q(m – 1)(n + r – 1)z = 0. (5.2) The equation of the line 15 26 is r(1 – n)(l + p – 1)x – (l(n(p + r – 1) + (p – 1)(r – 1)) + (n – 1)(p – 1)(r – 1))y + l(1 – p)(n + r – 1)z = 0.
(5.3)
It may now be verified that these three lines are concurrent at a point P if, and only if, equation (2.8) holds. The co-ordinates of P (with r given by (2.8)) are (x, y, z), where x = – l(m(n(p(q + r – 1) + q(r – 1)) + q(p – 1)(r – 1)) + n(p(q(r – 2) – r + 1) + (1 – r)(2q – 1)) + q(1 – r)(p – 1)) + mq(1 – r)(1 – n)(1 – p) + q(1 – n)(1 – p)(1 – r), (5.4) 4
y = l(r(n(p(q – 1) – 2q + 1) + q) – m(n((p(q – r) – q(r + 1) + r) + q(p(r – 1) + 1))) + qr(m – 1)(n – 1)(p – 1), z = l(m(n(p(q – r) + q(r – 1)) + q(p – 1)(r – 1)) + npr(1 – q)) + mq(n – 1)(p – 1)(r – 1).
(5.5) (5.6)
It now follows by symmetry that lines 13 24, 46 35, 61 52 are concurrent at a point Q and that lines 15 26, 35 46, 14 23 are concurrent at a point R. Finally it may be shown that AP, BQ, CR meet at a point Y with co-ordinates (x, y, z), where x = {p(m – 1)}/{m(n – p) – p(n – 1)}, (5.7) y = {l(q – 1)}/{l(n – q) – q(n – 1)}, (5.8) z = {p(1 – q)}/{n(p + q – 1) – pq}. (5.9)
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2011/160
Tangents to a Conic from the vertices of a Triangle Christopher J Bradley
L1 36 P
L2
13 26 A
F 2 3 23
1 Y
16 6
Z
E
X L4
15
24
L5
C
B 4
45 5
R 46 14
D 35 Q 25
L3
L6
Fig. 1 Tangents from A, B, C to Conic 123456 Abstract: Given a conic which cuts the sides of a triangle in six real points, tangents A1, A2, B3, B4, C5, C6 are drawn to the conic. The tangents meet each other at 12 points (15 if you count A, B, C). These form a 12 point configuration (see Article 159) so that six of the twelve points lie
1
on a conic and the other six form 3 lines concurrent at a point. Further properties of the configuration emerge. 1. Introduction In Fig. 1 the six tangents are labelled L1, L2, L3, L4, L5, L6. L1^L2 = A, L3^L4 = B, L5^L6 = C. The other 12 intersection points are labelled ij = Li^Lj. It is proved that 13, 26, 15, 46, 35, 24 lie on a conic and that lines 36 45, 25 16, 14 23 are concurrent at a point X. Several other concurrences are established. For example 36 45, 26 15, 24 13 are concurrent at P, 35 46, 24 13, 16 25 at Q, 14 23, 26 15, 35 46 at R, and PA, QB, CR at Y. It is also established that if lines 3 4, 5 6 meet at D then Q, 35, D, 46, 14 are collinear, with similar results for E, F. Finally AD, BE, CF are concurrent at a point Z. 2. Choice of Conic, the co-ordinates of points and equations of lines We choose the initial conic to have equation y2 = zx. Points 1, 2, 3, 4, 5, 6 are given parameters p, q, r, t, u, v respectively. Point K on this conic obviously may be taken to have co-ordinates (1, k, k2). So 1, for example has co-ordinates (1, p, p2). The tangents at K will have equation 2yk = k2x + z. (2.1) A = 1^2 being the intersection of the tangents at 1 and 2 (with parameters p and q) has coordinates A(2, p + q, 2pq). Other points such as 35 (parameters r and u) have co-ordinates 35(2, r + u, 2ru). 3. The conic through 13, 26, 15, 46, 35, 24 The conic has the form ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where a = 2(p(q(r(t(u – v) + uv) – tuv) + rtuv) – qrtuv), (3.2) b = 8(pru – qtv), (3.3) c = 2(p – q + r – t + u – v), (3.4) f = – 2(p(r + u) – q(t + v) + ru – tv), (3.5) g = p(q(r – t + u – v) + r(t – u + v) + t(u – v) + uv) – q(r(t – u + v) + t(u – v) + uv) + r(t(u – v) + uv) – tuv, (3.6) h = 2(qtv(r + u) – p(q(ru – tv) + ru(t + v))) (3.7) This may be checked by showing that the co-ordinates of the six points lie on the conic with this equation.
2
4. The points X, P, Q, R, Y The line 36 45 has equation (r(t(u – v) – uv) + tuv)x + 2(rv – tu)y – (r – t – u + v)z = 0.
(4.1)
The line 23 14 has equation (p(q(r – t) – rt) + qrt)x + 2(pt – qr)y – (p – q – r + t)z = 0.
(4.2)
The line 25 16 has equation (p(q(u – v) – uv) + quv)x + 2(pv – qu)y – (p – q – u + v)z = 0.
(4.3)
It may be checked that these lines all pass through the point X with co-ordinates (x, y, z), where x = –2(p(t – v) – q(u – r) + rv – tu), (4.4) y = p(q(r – t – u + v) – rt + uv) + q(rt – uv) + r(t(u – v) – uv) + tuv, (4.5) z = 2(p(q(rv – tu) – tv(r – u)) + qru(t – v)). (4.6) It is the case that P = 36 45 ^ 15 26 ^ 24 13. The equation of 36 45 is given in equation (4.1). The equation of 15 26 is (quv – p(q(u – v) + uv))x + 2(pu – qv)y – (p – q + u – v)z = 0. (4.7) And the line 24 13 has equation (p(q(r – t) + rt) – qrt)x + 2(qt – pr)y + (p – q + r – t)z = 0.
(4.8)
It may be checked that these lines all pass through P with co-ordinates (x, y, z), where x= 2(p(r – u) + q(v – t) – rv + tu), y = p(q(r – t – u + v) + rt – uv) + q(uv – rt) + r(t(u – v) – uv) + tuv, z = 2(p(q(rv – tu) + ru(t – v)) + qtv(u – r)).
(4.9) (4.10) (4.11)
The point Q = 24 13 ^ 35 46 ^ 25 16. The equation of 24 13 is given by (4.8) and that of 25 16 is given by (4.3). The equation of 35 46 is (tuv – r(t(u – v) + uv))x + 2(ru – tv)y – (r – t + u – v)z = 0. (4.12) It may now be checked that Q has co-ordinates (x, y, z), where x = 2(p(r – v) + q(u – t) – ru + tv), y = p(q(r – t + u – v) + rt – uv) + q(uv – rt) – r(t(u – v) + uv) + tuv, z = 2(p(q(ru – tv) + rv(t – u)) + qtu(v – r)).
(4.13) (4.14) (4.15)
The point R = 35 46 ^ 15 26 ^ 23 14. The equation of 35 46 is (4.12), the equation of 15 26 is (4.3) and the equation of 23 14 is (4.2). It may now be checked that R has co-ordinates (x, y, z), where 3
x = – 2(p(t – u) + q(v – r) + ru – tv), y = p(q(r – t + u – v) – rt + uv) + q(rt – uv) – r(t(u – v) + uv) + tuv, z = 2(p(q(ru – tv) – tu(r – v)) + qrv(t – u)).
(4.16) (4.17) (4.18)
The fact that AP, BQ, CR are concurrent at a point Y is left to the reader to check .The equations of the lines and the co-ordinates of Y are extremely lengthy. 5. The points D, E, F and Z = AD ^ BE ^ CF The lines 3 4 and 5 6 are respectively rtx – (r + t)y + z = 0 and uvx – (u + v)y + z = 0.
(5.1)
These lines meet at D with co-ordinates (r + t – u – v, rt – uv, r(t(u + v) – uv) – tuv).
(5.2)
Similarly E has co-ordinates (p + q – u – v, pq – uv, p(q(u + v) – uv) – quv), and F has co-ordinates (p + q – r – t), pq – rt, p(q(r + t) – rt) – qrt).
(5.3)
The fact that AD, BE, CF are concurrent at a point Z is left to the reader to check .The equations of the lines and the co-ordinates of Z are extremely lengthy. Finally it may be shown that D lies on the line 35 46 as does Q, so that D, Q, R, 35, 46 are collinear as are E, R, P, 15, 26 and F, P, Q, 13, 24.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article: CJB/2011/161
When 12 points display eight 6-point Conics and 6 Concurrent Lines Christopher J Bradley Z
L6 L5
Q 24
36 26
A
23
L2
25
E
F
R X
35
P
L1 46
13 C 15 B
Q'
D 14
L3
45
L4
To Y
16
Fig. 1 Six tangents to a circle produce 12 points with surprising properties Abstract: Six lines being tangents to a circle at vertices of triangles in perspective create twelve points of intersection. Six conics with six points each can be drawn through the twelve points, each point lying on three of the conics. Also six lines can be draw through the twelve points, lines which are concurrent at the vertex of perspective. 1. Introduction Triangles ABC and DEF are inscribed in a circle and AD, BE, CF are concurrent at an arbitrary point P. Tangents at A and D meet at X. Tangents at B and E meet at Y. Tangents at C and F 1
Y
meet at Z. The line XYZ is the polar of P. Points L1, L2, L3, L4, L5, L6 are the labels given to the tangents at D, A, E, B, C, F respectively. The point 13 is the intersection of L1 and L3. Points 14, 15, 16, 23, 24, 25, 26, 35, 36, 45, 46 are similarly defined points of intersection. It is proved that the following sets of six points lie on a conic: (i) 24 36 25 13 45 16, (ii) 26 23 35, 15, 14, 46, (iii) 24 26 35 15 46 13, (iv) 16 46 23 35 14 25, (v) 16, 14, 45, 36, 23, 25, (vi) 24, 26, 36, 15, 13, 45, (vii) 26 36 23 15 45 14, (viii) 24 46 16 25 35 13. It is also shown that the following six lines all pass through the vertex of perspective P: (i) 13 24, (ii) 14, 23, (iii) 15 26, (iv) 16 25, (v) 35 46, (vi) 36 45. A number of other properties hold. For example lines BC, EF, 46 35 are concurrent at a point R on XYZ and two other such cases. And also line CF and conics (i) and (iii) have two points Q, Q' in common and two other such cases. Proofs of the above results are given using areal co-ordinates with ABC as triangle of reference. 2. Points D, E, F and the six tangents L1 – L6 and points X, Y, Z We take P to have co-ordinates (p, q, r).. The circle through A, B, C has, of course, the equation a2yz + b2zx + c2xy = 0. (2.1) D is the point where AP meets the circle again and so has co-ordinates D(– a2qr, q(b2r + c2q)), r(b2r + c2q)). Similarly E, F have co-ordinates E(p(c2p + a2r), – b2rp, r(c2p + a2r)), F(p(a2q + b2p), q(a2q + b2p), – c2pq). The tangents at A, B, C labelled L2, L4, L5 respectively have equations c2y + b2z = 0, a2z + c2x = 0, b2x + a2y = 0. The tangents at D, E, F labelled L1, L3, L6 respectively have equations (b2r + c2q)2x + a2b2r2y + c2a2q2z = 0, a2b2r2x + (a2r + c2p)2y + b2c2p2z = 0, c2a2q2x + b2c2p2y + (a2q + b2p)2z = 0. L1 and L2 meet at X with co-ordinates (a2(b2r – c2q), – b2(b2r + c2q), c2(b2r + c2q)). L3 and L4 meet at Y with co-ordinates (a2(a2r + c2p), b2(c2p – a2r), – c2(a2r + c2p)). L5 and L6 meet at Z with co-ordinates (– a2(a2q + b2p), b2(a2q + b2p), c2(a2q – b2p)). 3. The co-ordinates of the twelve points
2
(2.2)
(2.3)
(2.4) (2.5) (2.6)
We remind the reader that point ij is the intersection of Li and Lj. The co-ordinates of the twelve points are: 13: x = a2(b2rp – c2pq – a2qr), y = b2(a2qr – b2rp – c2pq), z = a2r(2b2r + c2q) + c2p(b2r + c2q). (3.1) 2 2 2 2 14: x = – a r, y = b r + 2c q, z = c r. (3.2) 2 2 2 2 15: x = – a q, y = b q, z = 2b r + c q. (3.3) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 16: x = – a (a qr + b rp – c pq), y = a q(b r + 2c q) + b p(b r + c q), z = c (a qr – b rp – c pq). (3.4) 2 2 2 2 23: x = a r + 2c p, y = – b r, z = c r. (3.5) 2 2 2 24: x = a , y = b , z = – c . (An ex-symmedian point) (3.6) 2 2 2 25: x = a , y = – b , z = c . (An ex-symmedian point) (3.7) 2 2 2 2 26: x = a q + 2b p, y = b , z = – c q. (3.8) 2 2 2 2 35: x = a p, y = – b p, z = 2a r + c p. (3.9) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 36: x = – a (a qr + b rp + c pq) – 2b c p , y = b (a qr + b rp – c pq), z = c (a qr – b rp + c2pq). (3.10) 2 2 2 45: x = – a , y = b , z = c . (An ex-symmedian point) (3.11) 2 2 2 2 46: x = a p, y = 2a q + b p, z = – c p. (3.12) 4. The eight conics The equation of a conic, using areal co-ordinates, is ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(4.1)
We give the values of the variables u, v, w, f, g, h for each of the six conics. Conic 24 36 25 13 45 16: u = 2a2b2c2qr (b2r + c2q), v = 2a2b2c2rp (b2r + c2q), w = 2a2b2c2pq (a2q + b2p). f = a2(a4qr(b2r + c2q) + a2p(b4r2 + c4q2) + b2c2p2(b2r + c2q)), g = b2(a4qr(b2r + c2q) + a2p(b4r2 + c4q2) + b2c2p2(b2r + c2q)), h = c2(a4qr(b2r + c2q) + a2p(b4r2 + c4q2) + b2c2p2(b2r + c2q)).
(4.2)
Conic 26 23 35 15 14 46 u = 2a4b2c2q2r2 (b2r + c2q), v = 2a2b4c2r2p2 (a2r + c2p), w = 2a2b2c4p2q2 (a2q + b2p). f = a2 (a6q2r2 (b2r + c2q) + 2a4pqr (b4r2 + 2b2c2qr + c4q2) + a2p2 (b6r3 + 4b4c2qr2 + 4b2c4q2r + c6q3) + b2c2p3 (b4r2 + c4q2)), g = b2 (a6q2r2 (b2r + c2q) + 2a4b2pqr2 (b2r + 2c2q) + a2p2 (b2r + c2q) (b4r2 + 3b2c2qr + c4q2) + b2c2p3 (b2r + c2q) 2), h = c2 (a6q2r2 (b2r + c2q) + 2a4c2pq2r (2b2r + c2q) +a2p2 (b2r + c2q) (b4r2 + 3b2c2qr + c4q2) + b2c2p3 (b2r + c2q) 2). (4.3) Conic 24 26 35 15 46 13 3
u = – 2a2b2c2qr (b2r + c2q), v = – 2a2b2c2rp (a2r + c2p), w = 2a2b2c2pq (a2q + b2p). f = a2 (a4qr (b2r + c2q) + a2p (b2r + c2q)2 – b2c2p2(b2r – c2q)), g = b2 (a4qr (b2r – c2q) + a2p (b2r + c2q) 2 + b2c2p2 (b2r + c2q)), h = c2 (a4qr (3b2r + c2q) + a2p (3b2r + c2q) (b2r + c2q) + b2c2p2 (b2r + c2q)).
(4.4)
Conic 16 46 23 35 14 25 u = 2a2b2c2qr (b2r + c2q), v = – 2a2b2c2rp (a2r + c2p), w = 2a2b2c2pq (a2q + b2p). f = – a2 (a4qr (b2r + c2q) + a2p (b2r + c2q) 2 + b2c2p2 (b2r – c2q)), g = – b2 (a4qr (b2r + 3c2q) + a2p (b2r + 3c2q) (b2r + c2q) + b2c2p2 (b2r + c2q)), h = c2 (a4qr (b2r – c2q) – a2p (b2r + c2q) 2 – b2c2p2 (b2r + c2q)),
(4.5)
Conic 16 14 45 36 23 25 u = 2a2b4r2 (b2r + c2q), v = 2a4b2r2 (a2r + c2p), w = 2a2b2 (a4q2r + a2pq (2b2r + c2q) + b2p2 (b2r + c2q)), f = – a2 (a4qr (b2r – c2q) + a2p (b2r + c2q) (b2r – c2q) – b2c2p2 (b2r + c2q)), g = – b2 (a4qr (b2r – c2q) + a2p (b2r + c2q) (b2r – c2q) – b2c2p2 (b2r + c2q)), h = a4r(2b4r2 + b2c2qr + c4q2) + c2a2p(b2r + c2q)2 + b2c4p2(b2r + c2q).
(4.6)
Conic 24 26 36 15 13 45 u = 2a2c4q2 (b2r + c2q), v = 2c2a2 (a4qr2 + a2rp (b2r + 2c2q) + c2p2 (b2r + c2q)), w = 2a4c2q2 (a2q + b2p). f = a2 (a4qr (b2r – c2q) + a2p (b2r + c2q) (b2r – c2q) + b2c2p2 (b2r + c2q)), g = a4q (b4r2 + b2c2qr + 2c4q2) + a2b2p (b2r + c2q) 2 + b4c2p2 (b2r + c2q), h = c2 (a4qr (b2r – c2q) + a2p (b2r + c2q) (b2r – c2q) + b2c2p2 (b2r + c2q)).
(4.7)
Conic 25 35 13 24 46 16 u = 2b2c2 (a2qr + p (b2r + c2q)) (b2r + c2q), v = 2b2c4p2 (a2r + c2p), w = 2b4c2p2 (a2q + b2p), f = a6qr (b2r + c2q) + a4p (b2r + c2q) 2 + a2b2c2p2 (b2r + c2q) + 2b4c4p3, g = b2 (a4r (b2r + c2q) + a2p (b4r2 + c4q2) – b2c2p2 (b2r + c2q)), h = c2 (a4r (b2r + c2q) + a2p (b4r2 + c4q2) – b2c2p2 (b2r + c2q)). (4.8) Conic 26 36 23 15 45 14 u = – 2a2b2c2qr (b2r + c2q), v = 2a2b2c2rp (a2r + c2p), w = 2a2b2c2pq (a2q + b2p), f = – a2 (a4qr (b2r + c2q) + a2p (b4r2 + 4b2c2qr + c4q2) + 3b2c2p2 (b2r + c2q)), g = – b2 (a4qr (b2r + c2q) + a2p (b4r2 + 2b2c2qr – c4q2) + b2c2p2 (b2r + c2q)), h = – c2 (a4qr (b2r + c2q) – a2p (b4r2 – 2b2c2qr – c4q2) + b2c2p2 (b2r + c2q)).
5. The six concurrent lines
4
(4.9)
Their equations are: 13 24: (b2r + c2q)x – (a2r + c2p)y + (a2q – b2p)z = 0. 14 23: r(b2r + c2q)x + r(a2r + c2p)y – (a2qr + b2rp + 2c2pq)z = 0. 15 26: q(b2r + c2q)x – (a2qr + 2b2rp + c2pq)y + q(a2q + b2p)z = 0. 16 25: (b2r + c2q) x + (a2r – c2p) y – (a2q – b2p) z = 0. 35 46: (2a2qr + b2rp + c2pq)x – (a2rp + c2p2)y – (a2pq + b2p2)z = 0. 36 45: (b2r – c2q)x + (a2r + c2p)y – (a2q + b2p)z = 0.
(5.1) (5.2) (5.3) (5.4) (5.5) (5.6)
It may now be checked that all these six lines pass through the point P (p, q, r).
6. Further observations The polar line XYZ of P has equation (b2r + c2q) x + (a2r + c2p) y + (a2q + b2p) z = 0.
(6.1)
But note that not is not just the polar of P with respect to circle ABC but also to the eight conics of the system The line BC intersects XYZ at a point R (0, a2r + c2p, – a2q – b2p). It may be checked that R also lies on the line 35 46 (equation (5.5)) and also EF with equation (a2r + c2p) y + (a2q + b2p) z = a2qr (a2qr + b2rp + c2pq) x. (6.2) There are two further instances of this type involving CA and AB. Now FC has equation q (a2q + b2p) x = p (a2q + b2p) y.
(6.3)
It may now be shown that FC, conic 24 26 35 13 15 46 and conic 24 36 25 13 45 16 share two common points. See Fig. 1 for other instances of concurrences.
Flat 4, Terrill Court, 12-14, Apsley Road, 5
BRISTOL BS8 2SP.
6
Article: CJB/2011/162
When 24 Points form three 8-point Conics and 12 Concurrent Lines Christopher Bradley Abstract: A construction is described involving two chords of a conic that results in 24 points that form three 8-point Conics and 12 Concurrent Lines.
twuv
atdv
13 To 45
L1
58 L3 17 14 18 dtbv W
A
48
15
awcu
78 avdt
D
38
47 P 12
T
ctdv
To 16 25
C
67
36 26
23
56
34
68
37
U
B
L6
V
46
To 28
avct 24
27
btcv
L8
budw
L2
tuvw L4
35 btdv auwc L7 L5
Fig. 1 57 Three 8-point conics and 12 concurrent lines 1. Introduction In a given conic Σ two chords AC and BD are drawn intersecting at a point P. The tangents at A, B, C, D are labelled L1, L2, L3, L4 respectively. Lines Li, Lj intersect at a point labelled ij. It follows that points 13 and 24 lie on the polar p of P with respect to Σ. Line 12 34 meets Σ again at T and V and line 14 23 meets Σ again at W and U. It is proved that these two lines both pass through P. This result and some others in this article may be proved by Brianchon’s theorem. 1
28
(Private communication with G C Smith, University of Bath). The tangents at T, U, V, W are labelled L5, L6, L7, L8 respectively. It follows that points 57 and 68 both lie on p. Using the same definition of points as before we may now define points 12, 14, 15, 16, 17, 18 (all on L1), 23, 25, 26, 27, 28 (all on L2), 34, 35, 36, 37, 38 (all on L3), 45, 46, 47, 48 (all on L4), 56, 58 (both on L5), 67 and 78. The points thus labelled other than those on p are the 24 points of the title. The following are the main results proved in this article (i) Points 12, 58, 14, 78, 34, 67, 23, 56 lie on a conic; (ii) points 35, 27, 46, 38, 17, 45, 28, 16 lie on a conic; (iii) points 15, 18, 48, 47, 37, 36, 26, 25 lie on a conic; (iv) lines 15 37, 25 47, 26 48, 67 58, 12 34, 17 35, 56 78, 23 14, 38 16, 28 46, 27 45, and 18, 36 are concurrent at P; (v) line UW contains 4 more important points and p contains 8 more important points. These features are all shown in Fig. 1. In addition to the three 8-point conics listed Cabri II plus indicates that there are two more 8-point conics A, B, U, W, 56, 27, 17, 58 and C, D, U, W, 78, 45, 35, 67 and numerous 6-point conics. Proofs are presented in the projective plane using homogeneous projective co-ordinates. 2. The conic Σ, chords AC, BD and the tangents L1 – L4 and their intersections We suppose the conic Σ has equation y2 = zx, with A the point A(1, 0, 0), B the point B(1, 1, 1), C the point C(0, 0, 1) and the point P on AC having co-ordinates P(s, 0, t). Later for convenience we shall set s = (2uv) 2 and t = (u2 – v2)2 in order that s, t, and s + t have rational square roots. The equation of BP is tx + (s – t)y – sz = 0.
(2.1)
This meets the conic Σ at the point D with co-ordinates D(s2, – st, t2). The tangent to the conic at a general point (1, q, q2) (with parameter q) has equation 2qy = q2x + z. (2.2) The equations of the tangents at A, B, C, D may now be determined. L1 Tangent at A: q = 0, z = 0. L2 Tangent at B: q = 1, 2y = x + z. L3 Tangent at C: q infinite, x = 0. L4 Tangent at D: q = – t/s t2x + 2sty + s2z = 0.
(2.3) (2.4) (2.5) (2.6)
The co-ordinates of the six intersection of these lines one with another may now be obtained. 12: x = 2, y = 1, z = 0. 13: x = 0, y = 1, z = 0. 14: x = – 2s, y = t, z = 0. 2
23: x = 0, y = 1, z = 2. 24: x = – 2s, y = t – s, z = 2t. 34: x = 0, y = – s, z = 2t. The points 13 and 24 determine the polar line p, which therefore has equation tx + sz = 0.
(2.7)
3. The second quadrilateral TUVW and lines L5 – L8 The equation of the line 12 34 is tx – 2ty – sz = 0.
(3.1)
This line meets the conic at points we label T and V whose co-ordinates are T: x = 4u4, y = 2u2 (u2 –v2), z = (u2 – v2)2, V: 4v4, y = 2v2 (v2 – u2), z = (u2 – v2)2. Here the values of s, t in terms of u, v (section 2) have been used. The equation of the line 14 23 is tx + 2sy – sz = 0.
(3.2)
This line meets the conic at points we label U and W whose co-ordinates are U: x = 4u2v2, y = 2uv (u + v) 2, z = (u + v) 4, W: x = 4u2v2, y = – 2uv (u – v) 2, z = (u – v) 4. Here again the values of s, t in terms of u, v (section 2) have been used. It may also be checked that 12 34 and 14 23 pass through P. The equations of the tangents to the conic at the points T, U, V, W may now be determined. L5 Tangent at T (u2 – v2)2x – 4u2 (u2 – v2) y + 4u4z = 0. (3.3) 4 2 2 2 L6 Tangent at U (u + v) x – 4uv (u + v) y + 4u v z = 0. (3.4) 2 2 2 2 2 2 4 L7 Tangent at V (u – v ) x + 4v (u – v ) y + 4v z = 0. (3.5) 4 2 2 2 L8 Tangent at W (u – v) x + 4uv (u – v) y + 4u v z = 0. (3.6) In terms of u and v, the tangent at D, line L4 (see Equation (2.6)) also has equation (u2 – v2)4x + 8u2v2 (u2 – v2)2y + 16u4v4z = 0. 4. The 28 points of intersection, four of which lie on the polar line
3
(3.7)
The eight lines L1 – L8 create 28 points on intersection, 4 of which lie on the polar line p. Their co-ordinates may now be worked out and are: 12: x = 2, y = 1, z = 0. 13: x = 0, y = 1, z = 0. (On the polar line p.) 14: x = – 8u2v2, y = (u2 – v2)2, z = 0. 15: x = 4u2, y = u2 – v2, z = 0. 16: x = 4uv, y = (u + v) 2, z = 0. 17: x = 4v2, y = v2 – u2, z = 0. 18: x = – 4uv, y = (u – v) 2. 23: x = 0, y = 1, z = 2. 24: x = – 8u2v2, y = u4 – 6u2v2 + v4, z = 2(u2 – v2)2. (On the polar line p.) 25: x = 4u2, y = 3u2 – v2, z = 2(u2 – v2). 26: x = 4uv, y = u2 + 4uv + v2, z = 2(u + v) 2. 27: x = – 4v2, y = u2 – 3v2, z = 2(u2 – v2). 28: x = – 4uv, y = u2 – 4uv + v2, z = 2(u – v) 2. 34: x = 0, y = – 2u2v2, z = (u2 – v2)2 35: x = 0, y = u2, z = (u2 – v2). 36: x = 0, y = uv, z = (u + v) 2. 37: x = 0, y = v2, z = (v2 – u2). 38: x = 0, y = – uv, z = (u – v) 2. 45: x = – 8u4v2, y = u2 (u2 – v2) (u2 – 3v2), z = (u2 – v2)3. 46: x = – 8u3v3, y = uv(u + v)2(u2 – 4uv + v2), z = (u – v)2(u + v)4. 47: x = 8u2v4, y = v2 (v2 – u2)(3u2 – v2), z = (u2 – v2)3. 48: x = 8u3v3, y = – uv(u – v)2(u2 + 4uv + v2), z = (u + v)2(u – v)4. 56: x = 4u3v, y = u (u + v) (u2 + 2uv – v2), z = (u – v) (u + v) 3. 57: x = – 4u2v2, y = (u2 – v2)2, z = (u2 – v2)2. (On the polar line p.) 58: x = 4u3v, y = u (u2 – 2uv – v2)(v – u), z = (u + v)(v – u)3. 67: x = 4uv3, y = – v (u + v)(u2 – 2uv – v2), z = (v – u)(u + v)3. 68: x = – 4u2v2, y = – 4u2v2, z = (u2 – v2)2. (On the polar line p.) 78: x = 4uv3, y = v (v – u) (u2 + 2uv – v2), z = (u + v) (u – v) 3. 5. The three 8-point Conics The 24 points of Section 4, not on the polar line, lie on three 8-point conics, with each point lying on one of these conics. Their equations are of the form lx2 + my2 + nz2 + 2fyz + 2gzx + 2hxy = 0. (5.1) We now provide the constants l, m, n, f, g, h for each of these conics.
4
Conic 12 58 14 78 34 67 23 56 l = (u2 – v2)4, m = – 16u2v2 (u2 – v2)2, n = 16u4v4, f = 4u2v2 (u4 – 6u2v2 + v4), g = u8 + 14u4v4 + v8, h = – (u2 – v2)2(u4 – 6u2v2 + v4).
(5.2)
Conic 35 27 46 38 17 45 28 16 l = (u2 – v2)4, m = – 16uv3 (u + v) (u – v) 3, n = 16u4v4, f = 8u2v3 (u – v)(u2 – 2uv – v2), g = 2v (u – v) (u6 – u4v2 + 6u3v3 + u2v4 + 2uv5 – v6), h = 2v (u + v) (v – u) 3(u3 – u2v – 3uv2 – v3). (5.3) Conic 15 18 48 47 37 36 26 25 l = (u2 – v2)4, m = – 16u3v (u – v)(u + v)3, n = 16u4v4, f = 8u3v2(u3 – u2v – 3uv2 – v3), g = 2u (u + v) (u6 + 2u5v – u4v2 + 6u3v3 + u2v4 – v6), h = – 2u (u – v) 2(u + v) 3(u2 – 2uv – v2). (5.4) If one includes the vertices of the quadrangles there are other 8-point conics, for example Conic A B U W 56 27 17 58 with l = 0, m = 8v2 (u2 – v2), n = 8u2v2, f = – 12u2v2, g = – u4 + 6u2v2 + 3v4, h = (u2 – v2)4. (5.5) 6. The above 12 pairs of points create 12 lines all passing through P We remind the reader that P has co-ordinates P(s, 0, t), with s = (2uv) 2 and t = (u2 – v2)2. We now give the equations of the 12 lines. Line 23 14: Line 12 34: Line 15 37: Line 35 17: Line 25 47: Line 27 45: Line 26 48: Line 28 46: Line 58 67: Line 56 78: Line 38 16: Line 18 36:
tx + 2sy – sz = 0. (6.1) tx – 2ty – sz = 0. (6.2) 2 2 2 2 2 2 2 2 4u (u – v ) y + 4u v z – (u – v ) x = 0. (6.3) 2 2 2 2 2 2 2 2 (u – v ) x + 4v (u – v ) y – 4u v z = 0. (6.4) 2 2 2 2 2 2 2 2 2 2 4 2 2 4 (u + v )(u – v ) (v – 3u )x + 4u (u – v )(u – 2u v – 3v )y + 4u2v2 (u2 + v2) (3u2 – v2) z = 0. (6.5) 2 2 2 2 2 2 2 2 2 2 2 2 2 (u – v ) (u – 3v ) x + (u – v ) (3u – v ) y + 4u v (3v – u )z = 0. (6.6) 2 2 2 2 2 2 2 2 (u – v ) (u + 4uv + v ) x – 4uv (u + v) (u – 4uv + v ) y – 4u2v2 (u2 + 4uv + v2) z = 0. (6.7) 2 2 2 2 2 2 2 2 2 2 2 2 (u – v ) (u + v ) (u – 4uv + v ) x + 4uv (u – v) (u + v ) (u + 4uv + v ) y – 4u2v2 (u2 + v2) (u2 – 4uv + v2) z = 0. (6.8) (u2 – v2)2(u2 + v2)(u2 – 2uv – v2)x + 4uv(u2 – v2)(u4 + 2u3v + 2uv3 – v4)y – 4u2v2 (u2 + v2) (u2 – 2uv – v2) z = 0. (6.9) (u2 – v2)2(u2 + v2) (u2 + 2uv – v2) x – 4uv (u2 – v2) (u4 – 2u3 – 2uv3 – v4) y – 4u2v2 (u2 + v2) (u2 + 2uv – v2) z = 0.(6.10) (u2 – v2)2x – 4uv (u2 – v2) y – 4u2v2z = 0. (6.11) 2 2 2 2 2 2 (u – v ) x + 4uv (u + v) y – 4u v z = 0. (6.12) 5
It is straightforward to show that P lies on all of the above 12 lines. 7. The Quadrilaterals and Points on the line UW The co-ordinates of the vertices of the quadrilaterals ABCD and TUVW have been obtained in Sections 2 and 3, and these enable us to find four more points on the line UW. First the equation of UW is (u2 – v2)2x + 4u2v2 (2y – z) = 0. (7.1) We now give the equations of eight lines through the vertices. AT: (u2 – v2) y – 2u2z = 0. AV: (u2 – v2) y + 2v2z = 0. BT: (u2 – v2) x + (v2 – 3u2) y + 2u2z = 0. BV: (u2 – v2) x + (3v2 – u2) y – 2v2z = 0. CT: (u2 – v2) x – 2u2y = 0. CV: (u2 – v2) x – 2v2y = 0. DT: (u2 – v2)3x – 2u2 (u2 – v2) (u2 – 3v2) y – 8u4v2z = 0. DV: (u2 – v2)3x + 2v2 (u2 – v2) (3u2 – v2) y + 8u2v4z = 0.
(7.2) (7.3) (7.4) (7.5) (7.6) (7.7) (7.8) (7.9)
These 8 lines provide 4 intersections that lie on the line UW. Their co-ordinates are: atdv = AT^DV: x = 4u2v2(3u2 + v2), y = – 2u2(u2 – v2)2, z = – (u2 – v2)3. avdt = AV^DT: x = 4u2v2(u2 + 3v2), y = – 2v2(u2 – v2)2, z = (u2 – v2)3. btcv = BT^CV : x = 4u2v2, y = – 2u2(u2 – v2), z = – (u2 – v2)(3u2 + v2). bvct = BV^CT: x = 4u2v2, y = 2v2 (u2 – v2), z = (u2 – v2) (u2 + 3v2). It is left to the reader to determine in similar manner four further points lying on TV. 8. Ten more points on the polar line The polar line p was first introduced in equation (2.7). It is also the polar line of P with respect to each of the three conics in Section 5. We now give the co-ordinates of four more points on p (other than the four 13 24 57 68 already introduced). bvdt = BV^DT: x = 4u2v2(3v2 – u2), y = – 2v2(u2 – v2)(3u2 – v2), z = (u2 – v2)2(u2 – 3v2). btdv = BT^DV: x = 4u2v2(3u2 – v2), y = – 2u2(u2 – v2)(u2 – 3v2), z = – (u2 – v2)2(3u2 – v2). avct = AV^CT: x = 4u2v2, y = 2v2(u2 – v2), z = – (u2 – v2)2. atcv = AT^CV: x = – 4u2v2, y = 2u2(u2 – v2), z = (u2 – v2)2. We now give the equations of the four lines forming the quadrilateral ABCD. AB: y = z. BC: x = y. 6
(8.1) (8.2)
CD: DA:
tx + sy = 0. ty + sz = 0.
(8.3) (8.4)
We now give (again) the co-ordinates of 57 and 68. These are: 57 = AB^CD: x = s, y = – t, z = – t. 68 = AD^BC: x = s, y = s, z = – t. We now give the equations of the four lines forming the quadrilateral TUVW. TU: (u + v) 2(u2 – v2) x – 2u (u3 + 3u2v + uv2 – v3) y + 4u3vz = 0. UV: (u + v) 2(u2 – v2) x – 2v (u3 – u2v – uv2 – v3) y – 4uv3z = 0. VW: (u – v) 2 (u2 – v2) x + 2v (u3 + u2v – 3u2v + v3) y + 4uv3z = 0. WT: (u – v) 2 (u2 – v2) x – 2u (u3 – 3u2v + u2v + v3) y – 4u3vz = 0.
(8.5) (8.6) (8.7) (8.8)
We now give the co-ordinates of two more points on the polar line. tuvw = TU^VW: x = – 4u2v2(u2 + 2uv – v2), y = 2uv(u2 – v2)(u2 – 2uv – v2), z = (u2 – v2)2(u2 + 2uv – v2). twuv = WT^UV; x = = – 4u2v2(u2 – 2uv – v2), y = – 2uv(u2 – v2)(u2 + 2uv – v2), z = (u2 – v2)2(u2 – 2uv – v2). We now give the equations of eight more lines relating vertices of the two quadrilaterals. AU: (u + v) 2y – 2uvz = 0. BU: (u + v) 2x – (u2 + 4uv + v2) y + 2uvz = 0. CU: 2uv (u + v) 2x – 4u2v2y = 0. DU: (u + v)(u2 – v2)2x – 2uv(u + v)2(u2 – 4uv + v2)y – 8u3v3z = 0. AW: (u – v) 2y + 2uvz = 0. BW: (u – v) 2x – (u2 – 4uv + v2) y – 2uvz = 0. CW: (u – v) 2x + 2uvy = 0. DW: (u – v)2(u2 – v2)2x + 2uv(u – v)2(u2 + 4uv + v2)y + 8u3v3z = 0.
(8.9) (8.10) (8.11) (8.12) (8.13) (8.14) (8.15) (8.16)
Finally these lines provide four further intersections lying on the polar line. Their co-ordinates are: aucw = AU^CW: x = – 4u2v2, y = 2uv(u – v)2, z = (u2 – v2)2. budw = BU^DW: x = – 4u2v2(u2 + 4uv + v2), y = 2uv(u + v)2(u2 – 4uv + v2), z = (u2 – v2)2(u2 + 4uv + v2), cuaw = CU^AW: x = – 4u2v2, y = – 2uv(u + v)2, z = (u2 – v2)2. dubw = DU^BW: x = – 4u2v2(u2 – 4uv + v2), y = – 2uv(u – v)2(u2 + 4uv + v2), z = (u2 – v2)2(u2 – 4uv + v2). In conclusion it must be emphasized that although quadrilateral ABCD is arbitrary, TUVW is related to it by the construction outlined in Section 3, which explains the large number of 7
concurrences. Also the analysis is limited to positive values of s and t, the reason for the choice of u and v being to exclude square roots from the co-ordinates of many of the points.
Flat 4 Terrill Court, 12 – 14, Apsley Road, BRISTOL BS8 2SP.
8
Article: CJB/2011/163
A Theorem on the Complete Quadrilateral Christopher Bradley
A
F
E
R
Q P B
C
D
Fig. 1 Four Conics passing through a point R on a Diagonal Abstract: Given a complete quadrilateral ABCDEF and two points P and Q then it is wellknown that conics ABCPQ, AFEPQ, BFDPQ, and CDEPQ have a common point R. It is proved that when P, Q are the midpoints of the diagonals BE, CF respectively, then R lies on the third diagonal AD.
1
1. Introduction Given a complete quadrilateral ABCDEF and two points P and Q then it is well-known that conics ABCPQ, AEFPQ, BFDPQ and CDEPQ have a common point R. It is proved that when P, Q are the midpoints of the diagonals BE, CF respectively, then R lies on the third diagonal AD. In the proof below we use areal co-ordinates with ABC the triangle of reference and DEF a transversal with D on BC, E on CA and F on AB. 2. Co-ordinates of D, E, F, P, Q Take the transversal DEF to have the equation lx + my + nz = 0, then the co-ordinates of D, E, F are D(0, n, – m), E(– n, 0, l), F(m, – l, 0). The point P which is the midpoint of BE has co-ordinates P (– n, l – n, l) and the point Q which is the midpoint of CF has co-ordinates Q(m, – l, m – l). (It may be checked that if T is the midpoint of AD with co-ordinates T(n – m, n, – m), then P, Q, T are collinear.) 3. The conic ABCPQ and its intersection R with the line AD A conic in areal co-ordinates has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where ratios of the constants u, v, w, f, g, h are determined by substituting in Equation (3.1) the co-ordinates of the five points determining the conic. For the conic ABCPQ their values are u = 0, v = 0, w = 0, f = mn(lm + nl – mn), g = l(l – n)(lm + mn – nl), h = l(l – m)(mn + nl – lm). (3.2) The equation of AD is my + nz = 0 and the point R where the conic and AD intersect has coordinates R(m2n2, l2n(n – m), l2m(m – n)). 4. The conics AEFPQ, BFDPQ, and CDEPQ all pass through R This, of course, needs no further proof, but, for the record we give the values of u, v, w, f, g, h for the three conics. Conic AFEPQ: u = 0, v = 2lm (lm + mn – nl), w = 2nl (mn + nl – lm), 2
f = (lm + mn – nl)(mn + nl – lm), g = l2(mn + nl – lm), h = l2(lm + mn – nl).
(4.1)
Conic BFDPQ: u = 2l2(m – n)(lm + nl – mn), v = 0, w = 2mn2(lm – mn – nl), f = m2n((lm – mn – nl), g = n(lm + nl – mn)(lm – mn – nl), h = lm(m – n)(lm + nl – mn). (4.2) Conic CDEPQ: u = 2l2(m – n)(lm + nl – mn), v = 2m2n(lm + mn – nl), w = 0, f = mn2(lm + mn – nl), g = nl(m – n)(lm + nl – mn), h = m(lm + nl – mn)(lm + mn – nl). (4.3) It is now straightforward to show that these conics all pass through R.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/164
The Transversal of a Quadrilateral Christopher Bradley Abstract: A transversal LMNP intersects a convex quadrilateral ABCD with L on AB etc. The harmonic conjugates of L, M, N, P are L', M', N', P'. It is proved that L'P' and M'N' intersect on the diagonal BD and that L'M' intersects N'P' on the diagonal AC.
L' D A
P'
P L Y X N B
M M'
C N'
X'
Fig. 1 Intersection properties of harmonic conjugates 1. Introduction
1
It is well-known that when a transversal LMNP intersects a convex quadrilateral ABCD with L on AB, M on BC, N on CD and P on DA then Menelaus’ theorem generalizes so that (AL/LB)(BM/MC)(CN/ND)(DP/PA) = 1. (1.1) There is no obvious generalization of Ceva’s theorem, but the harmonic conjugates L' of L in AB, M' of M in BC, N' of N in CD and P' of P in DA do have some properties. These are (i) L'M' and N'P' intersect at a point X on AC and (ii) L'P' and M'N' intersect at a point Y on BD and (iii) both X and Y lie on the transversal LMNP. In this article these results are proved using homogeneous projective co-ordinates. 2. The points A, B, C, D, L, M, N, P, X and Y We suppose that A, B, C, D lie on the conic with equation fyz + gzx + hxy = 0
(2.1)
with points A, B, C, D having co-ordinates A(1, 0, 0), B(0, 1, 0), C(0, 0, 1) and D(– ft(1 – t), g(1 – t), ht).
(2.2)
The lines BC, CA, AB have equations x = 0, y = 0, z = 0 respectively It may also be shown that lines DA, DB, DC have equations: DA: g(1 – t)z – hty = 0, (2.3) DB: htx + ft(1 – t)z = 0, (2.4) DC: gx + fty = 0. (2.5) Suppose now that the transversal t = LMNP has equation lx + my + nz = 0.
(2.6)
Then the co-ordinates of L, M, N, P, X, Y are L = t^AB x = – m, y = l, z = 0, M = t^BC x = 0, y = – n, z = m, N = t^CD x = fnt, y = – gn, z = gm – flt, P = t^DA x = gm(t – 1) – hnt, y = gl(1 – t), z = hlt, X = t^CA x = n, y = 0, z = – l, Y = t^DB x = fm(1 – t), y = fl(t – 1) + hn, z = – hm.
(2.7) (2.8) (2.9) (2.10) (2.11) (2.12)
3. The harmonic conjugates and their properties The co-ordinates of the points L' and M' are immediate and are L' x = m, y = l, z = 0, M' x = 0, y = n, z = m. 2
(3.1) (3.2)
In order to work out the co-ordinates of N' and P' it is easier to work through the point X', the harmonic conjugate of X on CA and use properties of harmonic ranges, rather than get involved with a lengthy calculation. The co-ordinates of X' are X' x = n, y = 0, z = l. (3.3) We may now use the fact that X'N meets DA at P' and X'P meets CD at N'. The results are P' x = 2flt(t – 1) + gm(1 – t) + hnt, y = gl(1 – t), z = hlt, (3.4) N' x = fnt(t – 1), y = gn(1 – t), z = flt(t – 1) + gm(1 – t) + 2hnt. (3.5) The equation of L'P is hlx – hmy – (2fl(t – 1) + hn)z' = 0, and the equation of M'N' is (fl(t – 1) + 2hn)x + f(t – 1)(my – nz) = 0.
(3.6)
(3.7)
It may now be checked that the intersection of these lines is the point Y. Similarly L'M' meets N'P' at the point X.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article/CJB/165
A description of properties of Pascal’s hexagon Christopher Bradley 12
Abstract: A description is given of properties of Pascal’s hexagon, illustrating how 6 Pascal lines fall into two sets of 3 concurrent lines, defining 2 Steiner points.
To 12 X
Y 21 F U A
31 W
To 32 and N
V RS Q
E
P 13
L
23
M
T Z
D
B 32
C
Fig. 1 6 Pascal lines and 2 Steiner points P and T Description without analysis It should be mentioned at the outset that a calculation involving homogeneous projective coordinates is possible and has been carried out by the author, but it requires an algebraic software programme, and the co-ordinates of points are very elaborate and not worth recording. Given is a hexagon ABCDEF inscribed in a conic (shown as a circle in Fig. 1). When we write BDCEAF we mean the points X = FB^CE, Y = DC^AF, Z = BD^EA or in terms of numbers 1
123456 would mean points 61^34, 23^56, 45^12. Pascal’s theorem is that three such points are collinear. So for the arrangement of letters BDCEAF the points X, Y, Z are collinear and form a Pascal line. As the numbers may be rotated (123456 to 234561 etc.) and pairs such as 51 42 can be exchanged, there are therefore 60 Pascal lines in general. Sometimes symmetry properties of the hexagon reduce the number of Pascal lines to a factor of 60. In Fig. 1 we show the six Pascal lines obtained by keeping A, B, C fixed and permuting D, E, F. To be precise if DEF is replaced by EFD the Pascal line LMN is produced. If DEF is replaced by FDE the Pascal line UVW is produced. The Pascal lines XYZ, LMN, UVW are concurrent at the Steiner point T. If DEF is replaced by DFE the Pascal line 32 Q 23 is produced. If DEF is replaced by FED the Pascal line 31 R 13 is produced. If DEF is replaced by EDF the Pascal line 21 S 12 is produced. These Pascal lines are concurrent at the Steiner point P.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/2011/166
Analytic treatment of a Romanian problem
A
F D'
E P
Q P' F'
E' B
C D
Fig. 1 Abstract: A triangle ABC is given and a conic that cuts each of its sides in two real points. The tangents at these points are drawn providing a hexagon of tangents circumscribing the conic, whose vertices are labelled DF'ED'FE as in Fig.1. It is shown that AD, BE, CF are concurrent as are AD', BE', CF' and DD', EE', FF'. The positions of the resulting perspectives P, P' and Q are located. In the Romanian question the conic was the nine-point circle and then P lies on the hyperbola ABCKOi, where K is the symmedian point and Oi is the isotomic conjugate of the circumcentre O. The isotomic conjugate of P then lies on the diameter of the circumcircle joining the Steiner point and the Tarry point. It is also shown that in the general case AD', EF', FE' are concurrent as are BE', DF', FD' and CF', DE', ED'. Areal co-ordinates are used throughout with ABC the triangle of reference. 1
1. The conic and the points D, E, F and P The abstract may be regarded as an introduction as it describes the situation completely. We suppose (for simplicity but therefore lacking in generality) that the conic passes through the feet of two Cevian points with co-ordinates (l, m, n) and (p, q, r). For the Romanian problem these are the midpoints and the feet of the altitudes The equation of a conic Σ in areals is of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(1.1)
where the ratios of u, v, w, f, g, h may be determined by five points of the conic. We take the five points to be (0, m, n), (l, 0, n), (l, m, 0), (0, p, q) and (p, 0, r) with (p, q, 0) as a check. We find u = 2mnqr, v = 2nlrp, w = 2lmpq, f = – lp(mr + nq), g = – mq(np + lr), h = – nr(lq + mp). (1.2) The tangent to Σ at the point (d, e, f) is 2mnqrdx + 2nlrpey + 2lmpqfz – nr (lq + mp)(ex + dy) – lp(mr + nq)(fy + ez) – mq(np + lr)(dz + fx) = 0. (1.3) It follows that the tangents at the points (0, m, n) and (0, q, r) on BC have equations lp(mr – nq)(ny – mz) – mn(2lqr + mrp + npq)x = 0 and lp(nq – mr)(ry – qz) – qr(lmr + nlq + 2mnp)x = 0. These tangents meet at D with co-ordinates (x, y, z), where x = – lp(mr – nq)2, y = mq(lr(mr + 3nq) + np(3mr + nq)), z = nr(lq(3mr + nq) + mp(mr + 3nq)),
(1.4) (1.5)
(1.6)
The co-ordinates of E, F may be found from Equations (1.6) by cyclic change of x, y, z and l, m, n and p, q, r. The equations of AD, BE, CF may now be obtained and it may then be shown that they are concurrent at a point P with co-ordinates (x, y, z), where x = lp/{l2qr + 3lp(mr + nq) + mnp2}, (1.7) 2 2 y = mq/{m rp + 3mq(np + lr) + nlq }, (1.8) 2 2 z = nr/{n pq + 3nr(lq + mp) + lmr }. (1.9)
2
A pure method of solution follows from Chasles’ theorem that a triangle and its polar reciprocal are in perspective. Coolidge, J. L. A Treatise on Algebraic Plane Curves. New York: Dover, p. 33, 1959. 2. The points D', E', F' and P' Consider the tangents that meet at D'. The one from a point on AB has equation nqr(mp – lq)x + nrp(lq – mp)y – pq(2lmr + nlq + mnp) = 0.
(2.1)
The one from a point on CA has equation mnq(lr – np)x – nl(lqr + 2mrp + npq)y + lmq(np – lr)z = 0.
(2.2)
These lines meet at the point D' with co-ordinates (lp(3mr + nq), mq(lr – np), nr(mp – lq)).
(2.3)
Consider next the tangents that meet at E'. (We suppose the Cevians are drawn as in Fig.1) The one from the point on BC has equation lp(mr – nq)(ny – mz) – mn(2lqr + mrp + npq)x = 0. (2.4) The one from the point on AB is mnr(lq – mp)x + nlr(mp – lq)y – lm(lqr + mrp + 2npq)z = 0.
(2.5)
These lines meet at the point E' with co-ordinates (lp(nq – mr), – m(2mrp + lqr + npq), nr(lq – mp)).
(2.6)
Consider next the tangents that meet at F'. The one from the point on BC has equation lp(nq – mr)(ry – qz) – qr(lmr + nlq + 2mnp)x = 0.
(2.7)
The one from the point on CA has equation mqr(np – lr)x – rp(lmr + mnp + 2nlq)y + mpq(lr – np)z = 0.
(2.8)
The equations of AD', BE', CF' are now easily written down and it may then be checked that they all pass through the point P' with co-ordinates (lp(nq – mr), mq(np – lr), nr(lq – mp)). (2.9) The lack of cyclic symmetry is due to the fact that D', E', F' depend on the positions of (l, m, n) and (p, q, r). (This is not the case for the point P.) 3. The lines DD', EE', FF' meet at a point Q
3
Using the co-ordinates obtained in Sections 1 and 2 we find the equations of DD', EE', FF' are as follows: DD': nq(l2qr – mnp2)x – nlp(2lqr + mrp + npq)y – lpq(lmr + nlq + 2mnp)z = 0. (3.1) EE': mn(3lr + np)x – nl(lr – np)y – lm(lr + 3np)z = 0. (3.2) FF': qr(lq + 3mp)x – rp(3lq + mp)y – pq(mp – lq)z = 0. (3.3) The fact that these three lines are concurrent is a result of Brianchon’s theorem. Johnson, R. A. §387 in Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, p. 237, 1929. The point of concurrence Q is found to have co-ordinates (x, y, z), where x = – lp(l2qr(3mr + nq) + lp(m2r2 + 8mnqr – n2q2) + mnp2(3mr + nq)), y = – mq(l3qr2 + 3l2pr(mr + 2nq) + lnp2(6mr + nq) – mn2p3), z = nr(l3q2r – l2pq(6mr + nq) – 3lmp2(mr + 2nq) – m2np3).
(3.4)
There are, of course many other Brianchon points, but this is the only one we consider. 4. The position of P in the Balkan problem This is when Σ is the nine-point circle and then (l, m, n) = (1, 1, 1) and (p, q, r) = (1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)). The co-ordinates of P are now (1/(a2(b2 + c2 – a2) + b2c2), 1/(b2(c2 + a2 – b2) + c2a2), 1/(c2(a2 + b2 – c2) + c2a2)).
(4.1)
Cabri II plus and these co-ordinates show that P lies on the hyperbola ABCKOi, where, K is the symmedian point and Oi is the isotomic conjugate of the circumcentre, which has equation fyz + gzx + hxy = 0, (4.1) where f = a2(b2 – c2)(b4 + c4 + a2b2 + a2c2),
(4.2)
and g, h follow from f by cyclic change of a, b, c. In the above it has been proved that the isotomic conjugate of P lies on the line through the circumcentre, the third Brocard point and the Steiner and Tarry points. Gallatly, W. "The Steiner and Tarry Points." §143 in The Modern Geometry of the Triangle, 2nd ed. London: Hodgson, p. 102, 1913. Details are left to the reader. 5. AD', EF', FE' are concurrent as are BE', DF', FD' and CF', DE', ED'
4
Finally there are three other concurrencies as indicated above. We find that AD', EF', FE' are concurrent at a point with co-ordinates (lp(mr + 3nq), mq(np – lr), nr(lq – mp). The co-ordinates of the other points of concurrence may be found by cyclic change of x, y, z and l, m, n and p, q, r.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2011/167
Basic Properties of a Quadrangle possessing an Incircle Christopher Bradley
E
F J G S A
X L
U
V
H R
Z P
D
M K
Y C Q
B
I
Fig. 1 Quadrangle and Incircle with associated Conic and their Polar Line 1. Introduction Starting with a circle PQRS tangents are drawn at P, Q, R, S to produce a quadrangle ABCD with P on AB etc. The intersection of the diagonals AC and BD is the point labelled Z. Tangents 1
AB and CD intersect at E and tangents AD and BC meet at H. By construction it follows that EH is the polar of Z with respect to circle PQRS. Point X = AR^BS, point Y = CP^DQ. We prove that XYZ is a straight line and that Z lies on both PR and QS. We prove that lines DP, AC, BS are concurrent at a point L and that lines BR, AC, DQ are concurrent at a point M. We also show that lines AR, BD, CS are concurrent at a point U and that lines AQ, BD, CP are concurrent at V. It then follows that points X, Y, L, M, U, V lie on a conic, showing that every quadrangle with an incircle also possesses an interior conic Σ. It also turns out that EH is also the polar of Z with respect to Σ. Finally we show that there are five other significant points on the polar line. These are F = BS^DQ, J = PS^QR, G = DP^BR, K = AR^CP, I = SR^PQ. It is also shown that J lies on BD and I lies on AC. Proofs are given using areal co-ordinates with PQR as triangle of reference. 2. Points P, Q, R, S, A, B, C, D, Z Points P, Q, R form the triangle of reference, so we take S to be a general point on the circumcircle with equation a2yz + b2zx + c2xy = 0, (2.1) 2 2 2 and therefore having co-ordinates S(– a t(1 – t), b (1 – t), c t). In Fig. 1 we have chosen 1 < t < ∞, so that PQRS appear in that cyclic order. The equation of the tangent at P which is the line AB has equation c2y + b2z = 0.
(2.2)
The equation of the tangent at Q which is the line BC has equation a2z + c2x = 0.
(2.3)
The equation of the tangent at R which is the line CD has equation b2x + a2y = 0.
(2.4)
The equation of the tangent at S which is the line DA has equation x/a2 + t2y/b2 + (1 – t)2z/c2 = 0.
(2.5)
The co-ordinates of the vertices of the quadrangle are therefore: A: x = a2(2t – 1), y = – b2, z = c2. B: x = a2, y = b2, z = – c2. C: x = – a2, y = b2, z = c2. D: x = – a2(1 – t), y = b2(1 – t), z = c2(1 + t).
(2.6) (2.7) (2.8) (2.9)
2
The equation of AC is x/a2 + ty/b2 + (1 – t)z/c2 = 0,
(2.10)
x/a2 – ty/b2 + (1 – t)z/c2 = 0.
(2.11)
and the equation of BD is
The point Z = AC^BD therefore has co-ordinates Z(– a2(1 – t), 0, c2)
(2.12)
Line PR has equation y = 0 and line QS has equation c2x + a2(1 – t)z = 0.
(2.13)
It follows immediately that Z lies on both PR and QZ. 3. Points U, V, L, M, X, Y and the interior conic Σ passing through these points The equation of AQ is c2x + a2(1 – 2t)z = 0.
(3.1)
The point V is the intersection of this line with BD (equation (2.11)) and so has co-ordinates x = a2(2t – 1), y = b2, z = c2. (3.2) The equation of the line CP is b2z = c2y,
(3.3)
from which we observe that V lies on CP also. The line AR has equation b2x = a2(1 – 2t)y.
(3.4)
The point U is the intersection of this line with BD (equation (2.11)) and so has co-ordinates x = a2(1 – t)(1 – 2t), y = b2(1 – t), z = c2(3t – 1). (3.5) The line CS has equation (1 – 2t)x/a2 – t2y/b2 + (1 – t)2z/c2 = 0.
(3.6)
It may now be checked that U also lies on CS. The equation of BS is x/a2 = t2y/b2 + (t2 – 1)z/c2.
3
(3.7)
The point L is the intersection of this line with AC (equation (2.10)) and so has co-ordinates x = a2(2t2 – t – 1), y = b2(1 – t), z = c2(1 + t) (3.8) The equation of the line DP is b2(1 – t)z = c2(1 + t) y.
(3.9)
It is now obvious that L also lies on DP. The equation of the line BR is b2x = a2y.
(3.10)
The point M is the intersection of this line with AC (equation (2.10)) and so has co-ordinates x = a2(t – 1), y = b2(t – 1), z = c2(t + 1).
(3.11)
c2(1 + t)x + a2(1 – t)z = 0.
(3.12)
The equation of the line DQ is
It is now obvious that M lies also on DQ. The lines BS and AR (equations (3.7) and (3.4)) meet at the point X, which therefore has coordinates x = a2(t2 – 1)(2t – 1), y = b2(1 – t2), z = c2(t2 + 2t – 1). (3.13) The lines DQ (equation (3.12)) and CP (equation (3.3)) meet at the point Y, which therefore has co-ordinates x = a2(t – 1), y = b2( t + 1), z = c2(t + 1). (3.14) It may now be checked that XYZ is a straight line. The equation of the conic through U, V, L, M, X, Y is of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.15)
and after some computation we find u = 2b4c4 (1 + t), v = 2c4a4t (1 – t) 2, w = 2a4b4 (1 + t) (1 – t) 2, f = a4b2c2t1 – t) (2t – 3), g = a2b4c2 (1 – t) (3t + 2), h = a2b2c4t (t – 3).
(3.16)
4. Points on the polar line
4
The tangents at P and R meet at the point E with co-ordinates E (a2, – b2, c2). The tangents at Q and S meet at the point H with co-ordinates H (– a2t, b2 (2 – t), c2t). The line EH has equation x/a2 + ty/b2 + (t – 1)z/c2 = 0. (4.1) It is obvious that by construction that this line is the polar line of Z with respect to circle PQRS. It may also be checked that it is the polar line of Z with respect to the interior conic Σ. Other points lying on the polar line are F, G, I, J, K defined as follows: F = BS^QD with co-ordinates F(a2t(t – 1), – b2(t2 + t – t), c2t(1 + t)). (4.2) 2 2 2 G = DP^BR with co-ordinates (a (1 – t), b (1 – t), c (1 + t)). (4.3) 2 2 I = PQ^RS with co-ordinates (a t, – b , 0). (4.4) 2 2 J = PS^QR with co-ordinates (0, b (1 – t), c t). (4.5) 2 2 2 K = AR^CP with co-ordinates (a (1 – 2t), b , c ). (4.6)
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2011/168
Properties of a pair of Diametrically Opposite Triangles Christopher Bradley Q
H R
T
G K
V A
F U
L
E O S B D W
P
C
J X
N
M
Fig. 1 A pair of diametrically opposite Triangles creates many lines of interest Abstract: Starting from a triangle ABC and its circumcircle centre O, triangle DEF is such that AD, BE, CF pass through O. Tangents at the six points and the nine lines joining their vertices create a configuration that is explored in this article. 1. Introduction
1
Given a triangle ABC and its circumcircle Σ, centre O, lines AO, BO, CO meet Σ again at points D, E, F respectively. The tangent to Σ at A meets BC at L, with M and N similarly defined, forming the polar line LMN of triangle ABC. Similarly PQR is the polar line of triangle DEF. Lines LMN and PQR are obviously parallel. Lines AE, AF, BD, BF, CD, CE are also drawn. Intersections of these lines with each other and with the six tangents provide nine sets of three points that are collinear. These lines are shown in Fig. 1 and of the lines in the figure there are ten pairs of parallel lines (which are again obvious from the fact that ABC and DEF are related as described above). Areal co-ordinates are used with ABC as triangle of reference. 2. Triangles ABC, DEF and the six Tangents The line AO has equation c2(a2 + b2 – c2)y = b2(c2 + a2 – b2)z = 0. This meets the circumcircle Σ with equation a2yz + b2zx + c2xy = 0 at the point D with co-ordinates (x, y, z), where x = – (c2 + a2 – b2)(a2 + b2 – c2), y = 2b2 (c2 + a2 – b2), z = 2c2 (a2 + b2 – c2).
(2.1)
(2.2)
(2.3)
Point E and F have co-ordinates that may be written down from equations (2.3) by cyclic change of x, y, z and a, b, c. Similarly lines BO and CO may be written down from equation (2.1) by cyclic change of x, y, z and a, b, c. The tangent to the circumcircle at a point (d, e, f) is a2(fy + ez) + b2(dz + fx) + c2(ex + dy) = 0.
(2.4)
We can now write down the equations of the tangents at the six vertices: A: c2y + b2z = 0. B: a2z + c2x = 0. C: b2x + a2y = 0. D: 4a2b2c2x + c2(a2 + b2 – c2)2y + b2(c2 + a2 – b2)2z = 0. E: c2(a2 + b2 – c2)2x + 4a2b2c2y + a2(b2 + c2 – a2)2z = 0. F; b2(c2 + a2 – b2)2x + a2(b2 + c2 – a2)2y + 4a2b2c2z = 0.
(2.5) (2.6) (2.7) (2.8) (2.9) (2.10)
3. The sides of the triangles and the two polar lines LMN and PQR Sides BC, CA, AB, of course, have equations x = 0, y = 0, z = 0 respectively. 2
The side EF has equation (c2 + a2 – b2)(a2 + b2 – c2)x – 2a2(b2 + c2 – a2)(y + z) = 0.
(3.1)
The sides FD, DE may be written down from equation (3.1) by cyclic change of x, y, z and a, b, c. The tangent at A meets BC at L(0, b2, – c2). The tangent at B meets CA at M(– a2, 0, c2). The tangent at C meets AB at N(a2, – b2, 0). It is easy to see that L, M, N lie on the line with equation x/a2 + y/b2 + z/c2 = 0. (3.2) The tangent at D meets EF at the point P with co-ordinates (x, y, z), where x = 2a2(a2 – b2 – c2)(b2 – c2), y = b2(b4 + c4 – a4 – 2b2c2 + 4c2a2), z = – c2(b4 + c4 – a4 – 2b2c2 + 4a2b2).
(3.3)
The co-ordinates of Q, R may now be written down from equations (3.3) by cyclic change of x, y, z and a, b, c. The three points P, Q, R lie on the line with equation b2c2(3a4 + (b2 – c2)2)x + c2a2(3b4 + (c2 – a2)2)y + a2b2(3c4 + (a2 – b2)2)z = 0.
(3.4)
Lines LMN and PQR meet at a point on the line at infinity with co-ordinates (a2(b2 – c2), b2(c2 – a2), c2(a2 – b2)),
(3.5)
and are therefore parallel. This is, of course obvious since triangle ABC and DEF are diametrically opposite. 4. Lines LOP, MOQ, NOR It may now be shown that lines LP, MQ, NR are concurrent at O. The equation of LP is 2b2c2x – (b2 + c2 – a2)(c2y + b2z) = 0. (4.1) The equations of MQ and NR follow from equation (4.1) by cyclic change of x, y, z and a, b, c. The equation of AD is equation (2.1) and the equations of BE, CF follow from the equation of AD again by cyclic change of x, y, z and a, b, c. 5. Lines AE, AF, BD, BF, CD, CE and points of intersection with each other and with other lines already defined
3
Line AE has equation 2c2y + (b2 + c2 – a2)z = 0.
(5.1)
Point S = AE^LP has co-ordinates S: x = (a2 + b2 – c2)(b2 + c2 – a2), y = – 2b2(b2 + c2 – a2), z = 4b2c2.
(5.2)
Point G = AE^CF has co-ordinates G: x = a2(b2 + c2 – a2)2, y = b2(b2 + c2 – a2)(c2 + a2 – b2), z = – 2b2c2(c2 + a2 – b2).
(5.3)
Point K = AR^MQ has co-ordinates K: x = – 4c2a2, y = – (b2 + c2 – a2)(c2 + a2 – b2), z = 2c2(c2 + a2 – b2).
(5.4)
Line AF has equation (b2 + c2 – a2)y + 2b2z = 0.
(5.5)
Point U = AF^LP has co-ordinates U: x = (b2 + c2 – a2)(c2 + a2 – b2), y = 4b2c2, z = – 2c2(b2 + c2 – a2).
(5.6)
Point V = AF^NR has co-ordinates V: x = 4a2b2, y = – 2b2(a2 + b2 – c2), z = (a2 + b2 – c2)(b2 + c2 – a2).
(5.7)
Point H = AF^BE has co-ordinates H: x = a2(b2 + c2 – a2)2, y = – 2b2c2(a2 + b2 – c2), z = c2(a2 + b2 – c2) (b2 + c2 – a2).
(5.8)
Line BD had equation 2c2x + (c2 + a2 – b2)z = 0.
(5.9)
Point U = BD^AF has co-ordinates U: x = – (b2 + c2 – a2)(c2 + a2 – b2), y = – 4b2c2, z = 2c2(b2 + c2 – a2).
(5.10)
Point U also lies on LP. Line BF has equation (c2 + a2 – b2)x + 2a2z = 0.
(5.11)
Point K also lies on BF. Point T = AD^BF has co-ordinates T: x = – 2c2a2(a2 + b2 – c2), y = b2(c2 + a2 – b2)2, z = c2(a2 + b2 – c2) (c2 + a2 – b2).
(5.12)
Point J = NR^BF has co-ordinates J: x = – 2a2(a2 + b2 – c2), y = 4a2b2, z = (a2 + b2 – c2)(c2 + a2 – b2).
(5.13)
4
Line CD has equation 2b2x + (a2 + b2 – c2)y = 0. Point J also lies on CD. Point S also lies on CD. Point X = BE^CD has co-ordinates X: x = a2(a2 + b2 – c2) (b2 + c2 – a2), y = – 2a2b2(b2 + c2 – a2), z = c2(a2 + b2 – c2)2.
(5.14)
(5.15)
Line CE has equation (a2 + b2 – c2)x + 2a2y = 0.
(5.16)
Point V also lies on CE 6. 9 more lines Line HR has equation b2c2(a2 + b2 – c2)(c2 + a2 – b2)x + c2a2(a2 + b2 – c2)(b2 + c2 – a2)y + a2b2(a4 + 2a2(c2 – b2) + b4 + 2b2c2 – 3c4)z = 0.
(6.1)
Point T also lies on HR. Line PJ has equation (a4 + 2a2(b2 + c2) + (b2 – c2)2)x + 2a2(a2 + b2 – c2)y + 2a2(c2 + a2 – b2)z = 0. Point W = PJ^BD has co-ordinates W: x = 2a2(c2 + a2 – b2), y = – (a2 + b2 – c2) (c2 + a2 – b2), z = – 4c2a2.
(6.2)
(6.3)
Line NS has equation 4b2c2x + 4c2a2y + (b2 + c2 – a2)(c2 + a2 – b2)z = 0.
(6.4)
Point W also lies on NS Line LK has equation (a2 + b2 – c2)(c2 + a2 – b2)x + 4a2(c2y + b2z) = 0.
(6.5)
Point V also lies on LK. Line TG has equation b2c2(c2 + a2 – b2)x + c2a2(a2 + b2 – c2)y + a2b2(b2 + c2 – a2)z = 0.
(6.6)
Point X also lies on TG. 5
Line RK has equation 2c2(c2 + a2 – b2)x + 2c2(b2 + c2 – a2)y + (a4 + 2a2(c2 – b2) + (b2 + c2)2)z = 0.
(6.7)
Point U also lies on RK. Point W also lies on CE. Line QS has equation 2b2(a2 + b2 – c2)x + (a4 + 2a2(b2 – c2) + (b2 + c2)2)y + 2b2(b2 + c2 – a2)z = 0.
(6.8)
Point V also lies on QS. Line JM has equation 4b2c2x + (a2 + b2 – c2)(b2 + c2 – a2)y + 4a2b2z = 0.
(6.9)
Point U also lies on JM. Line GL has equation b2c2(a2 + b2 – c2)(c2 + a2 – b2)x + a2(b2 + c2 – a2)2(c2y + b2z) = 0.
(6.10)
Point H also lies on GL. 7. The 9 further pairs of parallel lines No proof is needed as they are obvious from the fact that triangles ABC and DEF are diametrically opposite. They are (i) BC and EF, (ii) CA and FD, (iii) AB and DE, (iv) BF and CE, (v)AE and BD, (vi) CD and AF, (vii) UKR and WNS, (viii) WJP and VKL, (ix) VQS and UJM.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
6
Article: CJB/2011/169
Two In-Perspective Triangles inscribed in a Conic Christopher Bradley
21 31 L M
N
A B 32 F O C
23 E
D
P
Q
13 R
12
Fig. 1 In-perspective Triangles inscribed in a Conic and Points and Lines of significance Abstract: Triangles ABC and DEF are inscribed in a conic and are in perspective through a point O. Polar lines LMN and PQR are drawn and six points of concurrency are shown to define a number of important collinearities. Two conics also emerge.
1. Setting the Scene We use areal co-ordinates with ABC as triangle of reference. We take the circumconic to have equation fyz + gzx + hxy = 0 (1.1) and the point O to have co-ordinates O(u, v, w). The equation of AO is wy = vz
(1.2)
Line AO meets the circumconic again at D with co-ordinates (x, y, z), where x = fvw, y = – v(gw + hv), z = – w(gw + hv).
(1.3)
Similarly E has co-ordinates x = – u(hu + fw), y = gwu, z = – w(hu + fw).
(1.4)
and F has co-ordinates x = – u(fv + gu), y = – v(fv + gu), z = huv.
(1.5)
The equations of the tangents at the vertices are: A: hy + gz = 0. B: fz + hx = 0. C: gx + fy = 0. D: (gw + hv)2x + fgw2y + hfv2z = 0. E: fgw2x + (hu + fw)2y + ghu2z = 0. F: hfv2x + ghu2y + (gu + fv)2z = 0.
(1.6) (1.7) (1.8) (1.9) (1.10) (1.11)
The equation of EF is fvwx – u(fw + hu)y – u(fv + gu)z = 0.
(1.12)
The equations of FD and DE may be obtained from equation (1.12) by cyclic change of x, y, z and f, g, h and u, v, w. The tangent at A meets BC at L (0, – g, h). The tangent at B meets CA at M (f, 0, – h) and the tangent at C meets AB at N (– f, g, 0). The tangent at D meets EF at P with co-ordinates (x, y, z), where x = fu(gw – hv), y = – hfv2 – gu(gw + hv), z = fgw2 + hu(gw + hv)
(1.13)
The co-ordinates of Q and R follow from equations (1.13) by cyclic change of x, y, z and f, g, h and u, v, w. 2. The polar lines The equation of LMN is x/f + y/g + z/h = 0. The equation of PQR is f(g2w2 + ghvw + h2v2)x + g(h2u2 + hfwu + f2w2)y + h(f2v2 + fguv + g2u2) z = 0. The equations of LP, MQ, NR are (gw + hv)x = u(hy + gz), (hu + fw)y = v(fz + hx), (fv + gu)z = w(gx + fy).
(2.1)
(2.2)
(2.3) (2.4) (2.5)
It may now be checked that O lies on LP, MQ and NR. 3. Six lines and six more points The equations of the lines joining vertices of the two triangles are: AE: (fw + hu)y + guz = 0, BF: (gu + fv)z + hvx = 0, CD: (hv + gw)x + fwy = 0, AF: huy + (fv + gu)z = 0, BD: fvz + (gw + hv)x = 0, CE: gwx + (hu + fw)y = 0.
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
Points of intersection of these lines have labels and co-ordinates as follows: CD^AE = 23 x = fgwu, y = – gu(gw + hv), z = (fw + hu)(gw + hv), AE^BF = 31 x = (fv + gu)(fw + hu), y = ghuv, z = – hv(fw + hu), BF^CD = 12 x = – fw(fv + gu), y = (fv + gu)(gw + hv), z = hfvw, CE^AF = 21 x = (fv + gu)(fw + hu), y = – gw(fv + gu), z = ghwu, AF^BD = 32 x = hfuv, y = (gw + hv)(gu + fv), z = – hu(gw + hv), BD^CE = 13 x = – fv(hu + fw), y = fgvw, z = (hu + fw)(hv + gw).
(3.7) (3.8) (3.9) (3.10) (3.11) (3.12)
4. Additional points of concurrency and collinearity It may now be shown that 32 and 23 both lie on LP and that 13 and 31both lie on MQ and that 21 and 12 both lie on NR.
The line 31 21 has equation fghvwx + (fv + gu)(fw + hu)(hy + gz) = 0.
(4.1)
It may now be shown that L lies on the line 31 21. Similarly M lies on line 12 32 and N lies on line 23 13. The equation of line 12 13 is f((g2w2 + ghuv + h2v2) + ghu(gw + hv))x + f(gw(fw + hu)y +hv(fv + gu))z = 0. (4.2) It may now be shown that P lies on the line 12 13. Similarly Q lies on line 23 21 and R lies on line 31 32.
21
31 ae L
N
M
A B cd
ce
32 F O 23 E D
P
Q 13 R
Fig.2 12
Caf bf bd
5. The two conics In Fig.2 we show, in addition to the material from Fig.1, two conics, one expected and the other new as far as we are concerned. The first and unexpected conic passes through the six points we have just been describing, that is 12 21 23 32 31 13. It has the form ax2 + by2 + cz2 + 2pyz + 2qzx + 2rxy = 0,
(5.1)
a = 2fghvw(gw + hv), b = 2fghwu(hu + fw), c = 2fghuv(fv + gu), p = fk, q = gk, r = hk,
(5.2) (5.3) (5.4) (5.5)
where we find
where k = f2gvw2 + f2hv2w + fg2u w2 + fh2uv2 + g2hu2w + gh2u2v + fghuvw.
(5.6)
We now define the six points where triangles ABC and DEF intersect. They are ae = BC^FD: x = 0, y = v(fv+ gu), z = gwu, cd = AB^EF : x = u(fw + hu), y = fvw, z = 0 , bf = CA^DE: x = huv, y = 0, z = w(gw + hv), af = BC^DE: x = 0, y = huv, z = w(fw + hu), bd = CA^EF: x = u(gu + fv), y = 0, z = fvw, ce = AB^FD: x = gwu, y = v(gw + hv), z = 0.
(5.7) (5.8) (5.9) (5.10) (5.11) (5.12)
These points, as expected lie on a conic of the form (5.1), with a = 2fv2w2(gw + hv), b = 2gw2u2(fw + hu), c = 2hu2v2(fv + gu), p = – uvw(f2vw + fu(gw + hv) + 2ghu2), q = – uvw (g2wu + gv(hu + fw) + 2hfv2), r = – uvw(h2uv + hw(fv + gu) + 2fgw2).
(5.13) (5.14) (5.15) (5.16) (5.17) (5.18)
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
Article: CJB/2011/170
The Brocard Conics Christopher Bradley
ab
To 1
N 31 ef
af F
32
fa
bf fb
A
K O
21 ae ea ec ce
2 E
ca
23 db D
B bd
12
fd
dc C cd
L
de
13 bc
3
M
Fig. 1 Abstract: Triangles ABC and DEF are inscribed in a circle and AD, BE, CF are concurrent at the symmedian point K. Sides of the two triangles are extended and the tangents at the vertices are drawn. The polar lines of the two triangles coincide and points of intersection of the sides and tangents are shown to produce in addition four six-point conics all with their centres on the Brocard axis OK, where O is the circumcentre of ABC. 1. Introduction 1
A triangle ABC, symmedian point K, is inscribed in a circle Σ centre O. Lines AK, BK, CK meet Σ again at D, E, F respectively. It follows that ABC and DEF are a pair of in-perspective triangles in the Brocard porism. Lines EF, BC meet at L, and M and N are similarly defined. The tangents to Σ are drawn at the six vertices and it is proved that the tangents at A and D pass through L. Similarly the tangents at B and E pass through M and the tangents at C and F pass through M. Point ab is the intersection of the tangents at A and B and point bc, ca, de, ef, fd are similarly defined. We show that these six points lie on a conic. The equations of the lines joining pairs of vertices are obtained. The following points are now obtained: 23 = CD^AE, 31 = AE^BF, 12 = BF^CD, 21 = CE^AF, 32 = AF^BD, 13 = BD^CE. It is then shown that these six points lie on a conic. It is then proved that AE, BD, FC are concurrent at a point 3 lying on the line LMN. Points 1 and 2 are similarly defined. The point af is defined as the intersection of the tangents at A and F. Points bf, bd, cd, ce, ae are similarly defined. It is the proved that these six points also lie on a conic. The six points of (internal) intersection of the two triangles are defined as follows: fa = AB^EF, fb = AB^FD, db = BC^FD, de = BC^DE, ec = CA^DE, ea = CA^EF. It is then proved that these six points also lie on a conic. Finally it is proved that the centres of these conics lie on the Brocard axis OK (the diameter of the 7-point circle). It must be mentioned also that the Brocard ellipse that touches internally all six sides of the two triangles (the inconic of the Brocard porism) also has its centre on OK, so we have a system of five conics altogether all with their centres on a line. See Fig. 1 for a diagram of all these results. 2. Points D, E, F, L, M, N The equation of the circumcircle Σ is well-known to have the equation a2yz + b2zx + c2xy = 0
(2.1)
Since the symmedian point K has co-ordinates (a2, b2, c2) the line AK has equation b2z = c2y. The point of intersection D of AK and Σ therefore has co-ordinates D(– a2, 2b2, 2c2). Similarly E and F have co-ordinates E(2a2, – b2, 2c2) and F(2a2, 2b2, – c2). The equation of EF is b2c2x = 2a2(c2y + b2z)
2
(2.2)
The side EF meets BC (x = 0) at L(0, – b2, c2). Similarly M and N have co-ordinates M(a2, 0, – c2) and N(– a2, b2, 0). The equation of LMN is thus x/a2 + y/b2 + z/c2 = 0.
(2.3)
The tangent to Σ at A has equation c2y + b2z = 0 and the tangent to Σ at D has equation 4b2c2x + c2a2y + a2b2z = 0.
(2.4)
(2.5)
Both these tangents pass through L. Points M and N also lie on corresponding tangents to Σ. It follows that LMN is the common polar line of the two triangles. (In fact LMN has the property that pairs of triangles in the Brocard porism are in triple reverse perspective with vertices on the line LMN.) Tangents to Σ at B, C and E, F may be written down from equations (2.4) and (2.5) by cyclic change of x, y, z and a, b, c. 3. Points ab, bc, ca, de, ef, fd and the conic on which they lie Point ab is the intersection of the tangents at A and B. It has co-ordinates ab(a2, b2, – c2). Similarly bc has co-ordinates bc(– a2, b2, c2) and ca has co-ordinates ca(a2, – b2, c2). These are, of course, the three ex-symmedian points of triangle ABC. Point de is the intersection of tangents to Σ at D and E and has co-ordinates de(a2, b2, – 5c2). Similarly ef has co-ordinates ef(– 5a2, b2, c2) and fd has co-ordinates fd(a2, – 5b2, c2). Since K is also the symmedian point of triangle DEF it follows that these three points are the ex-symmedian points of triangle DEF. It may now be shown that these six point lie on the conic Σ1 with equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where u = 2b4c4, v = 2c4a4, w = 2a4b4, f = 3a4b2c2, g = 3a2b4c2, h = 3a2b2c4.
(3.2)
4. Sides EF, FD, DE and joins of vertices AE, BF, CD, AF, BD, CE The equations of these nine lines are: EF: 2a2(c2y + b2z) – b2c2x = 0. (4.1) The equations of FD and DE now follow from (4.1) by cyclic change of x, y, z and a, b, c. 3
AE: 2c2y + b2z = 0. (4.2) The equations of BF and CD now follow from (4.2) by cyclic change of x, y, z and a, b, c. AF c2y + 2b2z = 0. (4.3) The equations of BD and CE now follow from (4.3) by cyclic change of x, y, z and a, b, c. 5. Points 23 = CD^AE, 31 = AE^BF, 12 = BF^CD, 21 = CE^AF, 32 = AF^BD, 13 = BD^CE and the conic on which they lie These six points have co-ordinates 23: x = a2, y = – 2b2, z = 4c2. (5.1) The co-ordinates of 31 and 12 follow from equations (5.1) by cyclic change of x, y, z and a, b, c. 21: x = 4a2, y = – 2b2, z = c2. (5.2) The co-ordinates of 32 and 13 follow from equations (5.2) by cyclic change of x, y, z and a, b, c. It may now be shown that these six points lie on a conic Σ2 with equation of the form (3.1) where u = 4b4c4, v = 4c4a4, w = 4a4b4, f = 7a4b2c2, g = 7a2b4c2, h = 7a2b2c4. (5.3) 6. The points 1, 2, 3 Lines AE and BD meet at the point 3 with co-ordinates (a2, b2, – 2c2). It may be shown that point 3 lies on both LMN and CF. Points 1 = AD^BF^CE and 2 = AF^BE^CD also lie on LMN. 7. Intersections of tangents at the vertices and the conic on which they lie Point af is the intersection of tangents to Σ at points A and F. It has co-ordinates (3a2, b2, – c2). Points bd, ce have co-ordinates that may be written down from those of af by cyclic change of x, y, z and a, b, c. Point bf is the intersection of tangents to Σ at points B and F. It has co-ordinates (a2, 3b2, – c2). Points cd, ae have co-ordinates that may be written down from those of bf by cyclic change of x, y, z and a, b, c. These six points lie on a conic Σ3 with equation of the form (3.1) where u = 2b4c4, v = 2c4a4, w = 2a4b4, f = 11a4b2c2, g = 11a2b4c2, h = 11a2b2c4.
(7.1)
8. Intersection of sides of triangles ABC and DEF and the conic on which they lie Point fa = AB^EF has co-ordinates (2a2, b2, 0). Points db = BC^FD and ec = CA^DE have coordinates which may be written down from those of fa by cyclic change of x, y, z and a, b, c.
4
Point fb = AB^FD has co-ordinates (a2, 2b2, 0). Points dc = BC^DE and ea = CA^EF have coordinates which may be written down from those of fb by cyclic change of x, y, z and a, b, c. These six points lie on a conic Σ4 with equation of the form (3.1) where u = 4b4c4, v = 4c4a4, w = 4a4b4, f = – 5a4b2c2, g = – 5a2b4c2, h = – 5a2b2c4
(8.1)
9. The Brocard axis and the centres of conics Σ1 - Σ4 The Brocard axis OK has equation b2c2(b2 – c2)x + c2a2(c2 – a2)y + a2b2(a2 – b2)z = 0…
(9.1)
A conic with equation (3.1) is known to have a centre with x – co-ordinate x = vw – gv – hw – f2 + fg + hf
(9.2)
with y- and z- co-ordinates following from (9.2) by cyclic change of u, v, w and f, g, h. It may now be shown that the conics Σ1, Σ2 , Σ3, Σ4 have x-co-ordinates Σ1: a2(3(b2 + c2) – 5a2), Σ2: a2(7(b2 + c2) – 11a2), Σ3: a2(11(b2 + c2 – 13a2), Σ4: a2(5(b2 + c2) – a2).
(9.3) (9.4) (9.5) (9.6)
It may now be checked that these four points all lie on the Brocard axis, as of course do the centres of the 7-point circle and the Brocard ellipse (that touches all six sides of the two triangles.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2011/171
A Conic through the feet of two Cevians may lead to a second Conic Christopher Bradley
Z
A
Y
X
W'
V
F
E
W R
P
V' Q
B
U
U'
C
T
S
D
Fig. 1 When an interior conic has intersection properties as shown then conic ABCDEF exists Abstract: Points P and Q are two Cevian points so that the feet U, V, W, U, V', W' carry a conic then the chords UV, WV', W'U' intersect as shown if and only if ABCDEF is a conic. Other properties (not proved in this article) hold as may be inferred from Fig. 1. 1. Introduction We start with a triangle ABC and Cevian points P and Q and with U, V, W the feet of the Cevian point P and U', V', W' the feet of the Cevian point Q. The construction in Fig. 1 shows the case when UV, WV' and W'U' are concurrent at a point R and it is shown that this is the case if and only if points A, B, C, D, E, F lie on a circumconic. Lines AD, BE, CF also concur at R. The tangents to this conic exhibit certain properties, which we do not investigate further as they are 1
well known in relation to a circumconic. We use areal co-ordinates with ABC as triangle of reference. 2. The conic UVWU'V'W' It is well known that if a conic is drawn through the feet of a Cevian point, then its three other intersections with the sides of the triangle concerned are also the feet of a second Cevian point (or the conic touches the sides of the triangle if the two points are coincident). The point here are distinct and are P(f, g, h) and Q(u, v, w). We next obtain the equation of the conic in terms of f, g, h, u, v, w. The feet of the Cevians have co-ordinates as follows: U(0, g, h), V(f, 0, h), W(f, g, 0), U'(0, v, w), V'(u, 0, w), W'(u, v, 0).
(2.1)
The conic through these six points has an equation of the form px2 + qy2 + rz2 + ayz + bzx + cxy = 0,
(2.2)
where we find p = ghvw, q = hfwu, r = fguv, a = – fu(gw + hv), b = – gv(hu + fw), c = – hw(fv + gu).
(2.3) (2.4)
3. The points D, E, F The equations of UW, U’V’, VW' are UW: ghx – hfy + fgz = 0, U'V': vwx + wuy – uvz = 0, VW': hvx – huy – fvz = 0.
(3.1) (3.2) (3.3)
UW and U'V' meet at D with co-ordinates D(fu(hv – gw), gv(fw + hu), hw(fv + gu)).
(3.4)
VW' and U'V' meet at E with co-ordinates E(u(fw + hu), v(hu – fw), 2hwu)
(3.5)
VW' and UW meet at F with co-ordinates F(f(fv + gu), 2fgv, h(fv – gu))
(3.6)
4. The conic ABCDEF
2
The six points A, B, C, D, E, F lie on a conic if and only if some relationship between f, g, h, u, v, w exists. To be precise they lie on a conic with equation ayz + bzx + cxy = 0 (4.1) where a = fu(fv + gu)(gw – hv), b = 2fgv2(fw + hu), c = hw(fv + gu)(gu – fv).
(4.2) (4.3) (4.4)
provided (f2vw + fu(3hv – gw) + ghu2)(f2vw + fu(3gw – hv) + ghu2) = 0.
(4.5)
5. The chords AD, BE, CF and the point R The equations of the chords AD, BE, CF are as follows: AD: gv(fw + hu)z – hw(fv + gu)y = 0, BE: 2hwux – u(fw + hu)z = 0, CF: f(fv + gu)y – 2fgvx = 0.
(5.1) (5.2) (5.3)
These chords are coincident (irrespective of condition (4.5)) at the point R with co-ordinates x = 1, y = 2gv/(fv + gu), z = 2hw/(fw + hu) (5.4) 6. Chords U'W', UV and WV' and when they meet at R The chords U'W', UV and WV' have equations: U'W': vwx – wuy + uvz = 0, UV: ghx + hfy – fgz = 0, WV': gwx – fwy – guz = 0. These meet at R if and only if either f2vw + fu(3hv – gw) + ghu2 = 0 or f2vw + fu(3gw – hv) + ghu2 = 0.
(6.1) (6.2) (6.3)
(6.4) (6.5)
The conclusion is that A, B, C, D, E, F is a conic if and only if the chords above are concurrent and if concurrent then at the point R and conversely.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article: CJB/2011/172
The Inner Circle of a Cyclic Quadrilateral Christopher Bradley
H D S
A T
U
W P
V
R
G O Q
B
C
J
Fig. 1 A cyclic quadrilateral, its midpoint rectangular hyperbola and inner circle Abstract: Given a cyclic quadrilateral ABCD with midpoints of sides P, Q, R, S and midpoints of diagonals U, V then it is well-known that a rectangular hyperbola passes through P, Q, R, S, U, V as well as O, the centre and H, J the other two diagonal points. We find the equation of this hyperbola and that of the inner circle OUVT, centre G. The midpoint W of UV is also the midpoint of PR and of QS. 1. Introduction Let ABCD be a cyclic quadrilateral in which the midpoints of AB, BC, CD, DA, AC, BD are respectively P, Q, R, S, U, V. A rectangular hyperbola Σ passes through these six points and also passes through the diagonal points T, H, J and the centre O of circle ABCD. We use Cartesian co-ordinates to obtain its equation in terms of the co-ordinates of A, B, C, D. We define the circle OUVT as the ‘Inner Circle’ and its equation is also obtained. 1
2. The rectangular hyperbola Σ We take the circle ABCD to be the unit circle with its centre the origin and co-ordinates of A given by x = (1 – a2)/(1 + a2), y = 2a/(1 + a2) (2.1) Points B, C, D have parameters b, c, d replacing a. In terms of the parameters the midpoint P of AB has co-ordinates x = (1 – a2b2)/{(1 + a2)(1 + b2)}, y = (a + b)(1 + ab)/{(1 + a2)(1 + b2)}.
(2.2)
with similar expressions for the co-ordinates of BC, CD, DA, AC, BD. It is now straightforward to obtain the equation of the rectangular hyperbola through these points, which is of the form sx2 + ty2 + 2uxy + 2gx + 2fy = 0, (2.3) with s = 2(abc + abd + acd + bcd – a – b – c – d), t = – 2(abc + abd + acd + bcd – a – b – c – d), u = 2(abcd – ab – ac – ad – bc – bd – cd + 1), f = 2(abcd – 1), g = (abc + abd + bcd + acd + a + b + c + d).
(2.4) (2.5) (2.6) (2.7) (2.8)
This rectangular hyperbola Σ clearly passes through the origin O. The diagonal point T = AC^BD has co-ordinates (x, y), where x = –(abc + acd – abd – bcd – a + b – c + d)/(abc + acd – abd – bcd + a – b + c – d), y = 2(ac – bd)/ (abc + acd – abd – bcd + a – b + c – d).
(2.9) (2.10)
It may now be checked that T lies on Σ, and hence also on H = AB^CD and J = AD^BC. 3. The inner circle OUVT It can now be shown that a circle passes through O, U, V, T and has equation of the form h(x2 + y2) + 2gx + 2fy = 0, (3.1) where h = 2(abc + acd – abd – bcd + a – b + c – d), g = abc – abd + acd – bcd – a + b – c + d, 2
(3.2) (3.3)
f = 2(bd – ac).
(3.4)
The centre of this circle may be seen to be the midpoint of OT.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/173
The Auxiliary Circles of a Cyclic Quadrilateral Christopher Bradley
G A
S D
U O
X
N
P
K
L T E
Y
V
R F C
Z
Q J B
M
H
Fig. 1 The Midpoint Circles and the Semi-Diagonal Point Circle
1
1. Introduction The construction of Fig. 1 proceeds as follows. First the cyclic quadrilateral ABCD is drawn. The midpoints of AB, BC, CD, DA, AC, BD are labelled P, Q, R, S, U, V, the centre of ABCD is the point O and the diagonal points are T = AC^BD, H = AB^DC and K = AD^BC. (These are all finite points.) The 10 point rectangular hyperbola P, Q, R, S, O, U, V, T, K, H is shown. We establish the existence of the three midpoint circles OUVT, OPRH, OQSK and also the SemiDiagonal point Circle LMNXYZ, where L, M, N are the midpoints of OT, OH, OK and X = OK^HT, Y = OH^KT and Z = OT^HK. Work is carried out using Cartesian co-ordinates with circle ABCD the unit circle centre O, the origin, 2. The Cartesian co-ordinates of A, B, C, D, T, H, K, P, Q, R, S, U, V The co-ordinates of A are (x, y), where x = (1 – a2)/(1 + a2). y = 2a/(1 + a2).
(2.1)
Co-ordinates of B, C, D are similar but with parameters b, c, d respectively. The midpoint P of AB therefore has co-ordinates (x, y), where x = (1 – a2b2)/{(1 + a2)(1 + b2)}, y = (1 + ab)(a + b)/ )/{(1 + a2)(1 + b2)}.
(2.2)
The midpoints Q, R, S, U, V of BC, CD, DA, AC, BD follow similarly with appropriate changes of parameters. The equation of AC is (1 – ac)x + (a + c)y = (1 + ac),
(2.3)
(1 – bd)x + (b + d)y = (1 + bd).
(2.4)
and that of BD is
These lines meet at T = AC^BD with co-ordinates (x, y), where x = – (abc + acd – abd – bcd – a + b – c + d)/(abc + acd – abd – bcd + a – b + c – d), y = 2(ac – bd)/(abc + acd – abd – bcd + a – b + c – d).
(2.5) (2.6)
Point K = AD^BC has co-ordinates that may be written down from equations (2.5) and (2.6) by exchange of c and d. Point H = AB^CD has co-ordinates that may be written down from equations (2.5) and (2.6) by exchange of c and b.
2
3. The three midpoint circles OUVT, OPRH, OQSK and their centres L, M, N We now determine the equations of the three midpoint circles analysing first the circle OUVT. Since it passes through O we suppose it has equation of the form h(x2 + y2) + 2gx + 2fy = 0. (3.1) Substituting in the co-ordinates of U, V we find h = 2(abc + acd – abd – bcd + a – b + c – d), g = (abc + acd – abd – bcd – a + b – c + d), f = 2(bd – ac).
(3.2) (3.3) (3.4)
It may now be verified that this circle passes through T. For the circle OPRH we may use the above equations but with the exchange of b and c and for circle OQSK with the exchange of c and d. The co-ordinates of the centre L of circle OUVT are therefore (x, y) where x = – (abc + acd – abd – bcd – a + b – c + d)/{2(abc + acd – abd – bcd + a – b + c – d)}, (3.5) y = – (bd – ac)/(abc + acd – abd – bcd + a – b + c – d)}. (3.6) For the centre M of circle OPRH we may use the above with the exchange of b and c and for centre N with the exchange of c and d. It may now be shown that LM and TH have the same slope equal to 2(ad – bc)/ (abc + acd – abd – bcd + a – b + c – d).
(3.7)
It follows that triangle LMN is homothetic with the diagonal point triangle THK through O and half the size. 4. The intersections of the midpoint circles and the circumcircle The circles OUVT and OPRH meet at O and a point X which lies on TH. The circles OUVT and OQSK meet at O and a point Y on TK. The circles OPRH and OQSK meet at O and a point Z on HK. Details are left to the reader. (The co-ordinates of X, Y, Z are technically very complicated.) The circle OQSK and the circle ABCD meet at points G and J which may be shown to lie on HT. The circle OPRH and the circle ABCD meet at points E and F which may be shown to lie on TK. The circle OUVT is internal to circle ABCD. Finally it may be shown that the six points L, M, N, X, Y, Z lie on a circle. Again details are left to the reader with the warning that its equation is so complicated that it is doubtful that it could 3
be transcribed without error. Fortunately no calculation is necessary as circle LMN is the ninepoint circle of the diagonal point triangle. (Private communication from David Monk).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
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Article 174/cjb/2011/174
A Cevian Circle leads to Perspective Triangles Christopher Bradley
A F R V W T
E J S
B
U
Q
P
C
D Fig. 1 ABC and DEF are in perspective
Abstract: A circle is drawn through U, V, W the feet of a Cevian point T. Three other points are created on the sides of triangle ABC enabling a second triangle DEF to be drawn. It is shown that triangles ABC and DEF are in perspective. 1. Preliminary analysis 1
The equation of a circle in areals is of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(1.1)
The circle we are looking at in this short article passes through the feet U, V, W of the Cevians of some arbitrary point T interior to triangle ABC. Suppose T has co-ordinates (f, g, h), where f, g, h are positive. Then U has co-ordinates (0, g, h), V has co-ordinates (f, 0, h) and W has coordinates (f, g, 0). When these sets of co-ordinates are substituted in (1.1) we get three equations for u, v, w with solutions u = a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(h + f))/{2f(f + g)(g + h)(h + f)}, (1.2) 2 2 2 v = – a gh(f + g)(h + f) + f(g + h)(b h(f + g) – c g(h + f))/{2g(f + g)(g + h)(h + f)}, (1.3) 2 2 2 w = – a gh(f + g)(h + f) – f(g + h)(b h(f + g) – c g(h + f)/{2h(f + g)(g + h)(h + f)}. (1.4) This circle meets BC at U(0, g, h) and P with co-ordinates (0, y, z), where y = – a2g2h(f2 + f(g + h) + gh) – 2hf(g + h)(b2h(f + g) – c2g(h + f)), z = – a2gh(f2 + f(g + h) + gh) + 2hf(g + h)(b2h(f + g) – c2g(h + f))).
(1.5) (1.6)
The circle meets CA at V(f, 0, h) and Q with co-ordinates (x, 0, z), where x = a2gh(f2 + f(g + h) + gh) + f(g + h)(b2h(f + g) – c2g(h + f)), z = f(g + h)(b2h(f + g) + c2g(h + f)) – a2gh(f2 + f(g + h) + gh).
(1.7) (1.8)
The circle meets AB at W(f, g, 0) and R with co-ordinates (x, y, 0), where x = a2gh(f2 + f(g + h) + gh) – f(g + h)(b2h(f + g) – c2g(h + f)), y = f(g + h)(b2h(f + g) + c2g(h + f)) – a2gh(f2 + f(g + h) + gh).
(1.9) (1.10)
2. The sides of the second triangle We can now compute the equations of the sides PQ, WU, RV of the second triangle. These are PQ: a2gh((f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(h + f)))x + (f(g + h)(b2h(f + g) – c2g(h + f)) – a2gh(f + g)(h + f))y + (a2gh(f + g)(h + f) + f(g + h)(b2h(f + g) – c2g(h + f)))z = 0, (2.1) WY:
f(hy – gz) – ghx = 0,
(2.2)
RV: h(a2gh(f + g)(h + f) – f(g + h) (b2h(f + g) + c2g(h + f)))x + h(a2gh(f + g)(h + f) – f(g + h) (b2h(f + g) – c2g(h + f)))y + f(f(g + h)(b2h(f + g) + c2g(f + h)) – a2gh(f + g)(h + f))z = 0. (2.3) 3. The vertices D, E, F of the second triangle Their co-ordinates are as follows: 2
PQ^WU = D: x = f(a2gh(f + g)(h + f)(g – h) – f(g + h)2(b2h(f + g) – c2g(h + f))), (3.1) y = g(h + f)(a2gh(f + g)(f – h) – f(g + h)(b2h(f + g) + c2g(f – h))), (3.2) z = h(f + g)(a2gh(f – g)(h + f) – f(g + h) b2h(f – g) + c2g(h + f))). (3.3) RV^PQ = E: x = 2h(h + f)(a2gh(f + g))(f – h) – f(g + h)(b2h(f + g) + c2g(f – h)))(a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) – c2g(f + h))), (3.4) 2 2 y = 2h(a gh(h + f)(f + g)(h + f) – f(g + h)(b h(f + g)(f – h) + c2g(h + f)2))(a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(h + f))), (3.5) 2 2 2 z = (2h(a gh(f + g)(h + f) – f(g + h)(b h(f + g) + c g(h + f))))(2h(a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) – c2g(h + f)))). (3.6) WU^RV = F: x = f(f + g)(a2gh(f – g)(h + f) – f(g + h)(b2h(f – g) + c2g(h + f))), (3.7) y = 2fg(a2gh(f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(h + f))), (3.8) 2 2 2 2 2 z = h(a gh(f + g) (h + f) – f(g + h)(b h(f + g) + c g(f – g)(h + f))). (3.9) 4. The point of perspective S It may now be shown that AD, BE, CF meet at a point S, which is therefore the vertex of perspective of the two triangles. Its co-ordinates are x= h(f + g)(h + f)(a2gh(f + g)(f – h) – f(g + h)(b2h(f + g) + c2g(f – h)))(a2gh(f – g)(f + h) – f(g + h)(b2h(f – g) + c2g(h + f))), (4.1) 2 2 2 2 y = 2gh(h + f)(a gh(f + g)(f – h) – f(g + h)(b h(f + g) + c g(f – h)))(a gh(f + g)(h + f) – f(g + h)(b2h(f + g) + c2g(h + f))), (4.2) 2 2 2 2 z = (2h(a gh(f + g)(h + f) – f(g + h)(b h(f + g) + c g(h + f))))(h(f + g)(a gh(f – g)(h + f) – f(g + h)(b2h(f – g) + c2g(h + f)))). (4.3) We might add a slightly non-mathematical comment. If AD meets BC at X, with Y and Z similarly defined then circle XYZ is astonishingly close to the incircle of ABC irrespective of the initial point T. CABRI II plus does of course indicate a difference but it is marginal. The point J in Fig. 1 is the centre of circle XYZ.
Flat 4, Terrill Court, 12-14, Apsley Road, 3
BRISTOL BS8 2SP.
4
Article: CJB/2011/175
When two Triangles intersect in Conics and in Circles Christopher Bradley Abstract: If P is internal to a triangle and points D, E, F are defined on AP, BP, CP respectively so that AD/AP = BE/BP = CF/CP then the intersections of triangles ABC and DEF lie on a conic. If P is the symmedian point K then the conic is a circle.
A
W
S
F
E
T K
V B
U
R
C
D
Fig. 1 The intersection points of a pair of related triangle 1. Introduction
1
In Fig. 1 ABC is a triangle with symmedian point K. D, E, F lie on AK, BK, CK and are such that AD/AK = BE/BK = CF/CK = t, where t = 2. The intersections of the two triangles lie on a circle. In this article we show first that when K is a general point then URVSWT is a conic and that when P is K this conic is a circle. In general when t is a positive number other than 2, these results still hold, but proofs are only given here in the case t = 2. 2. The points of intersection of the triangles ABC and DEF Let P have co-ordinates (f, g, h). Then, by definition D, E, F have co-ordinates D(2f – 1, 2g, 2h), E(2f, 2g – 1, 2h), F(2f, 2g, 2h – 1). The equation of EF is therefore (2g + 2h – 1)x = 2f(y + z). (2.1) Line EF meets AB, z = 0, at the point W with co-ordinates W(2f, 2g + 2h – 1, 0) and it meets CA at the point S with co-ordinates S(2f, 0, 2g + 2h – 1). By cyclic change of x, y, z and f, g, h we find that the other points of intersection are U(0, 2g, 2f + 2h – 1), T(2h + 2f – 1, 2g, 0), V( 2f + 2g – 1,0, 2h) and R(0, 2f + 2g – 1, 2h). 3. The conic URVSWT The equation of the conic is of the form ux2 + vy2 + wz2 + 2pyz + 2qzx + 2rxy = 0, where we find u = 4gh(2g + 2h – 1), v = 4hf(2h + 2f – 1), w = 4fg(2f + 2g – 1), and p = – f(4f2 + 4f(g + h – 1) + 2g(4h – 1) – 2h + 1) q = – g(4g2 + 4g(h + f – 1) + 2h(4f – 1) – 2f + 1). r = – h(4h2 + 4h(f + g – 1) + 2f(4g – 1) – 2g + 1).
(3.1)
(3.2)
4. When the conic is a circle CABRI II plus indicates that the only point when this conic is a circle is when P lies at the symmedian point and f = a2, g = b2, h = c2. Then the test for a circle holds, namely (v + w – 2p)/a2 = (w + u – 2q)/b2 = (u + v – 2r)/c2, their common value being 2(2a2 + 2b2 + 2c2 – 1)2. 5. Brief discussion
2
(4.1)
When t varies from 2 the same results hold, namely that URVSWT is a conic and that this conic is a circle when P lies at K. The circle obtained in the limit as t tends to zero is the circumcircle of triangle ABC and as t tends to 1 is the triplicate ratio circle. The set of circles as a whole forms the coaxal system of circles orthogonal to the Apollonius circles, with centres on the Brocard axis and with radical axis the Lemoine axis.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article/CJB/2011/176
Orthologic Triangles Christopher Bradley Abstract: Triangles ABC and DEF are orthologic if the perpendiculars from the vertex of one of them to the corresponding sides of the other are concurrent. The property is symmetric but not transitive.
A F E Y X B
C D Fig. 1 A pair of orthologic triangles
1. The vertices and sides of the two triangles The Cartesian co-ordinates of the vertices are taken to be: A(a, p), B(b, q), C(c, r), D(d, u), E(e, v), F(f, w).
(1.1)
The equations of the sides of triangle DEF are DE: (u – v)x + (e – d)y + dv – eu = 0, EF: (v – w)x + (f – e)y + ew – fv = 0, FD: (w – u)x + (d – f)y + fu – dw = 0.
(1.2) (1.3) (1.4)
2. Perpendiculars from A, B, C onto EF, FD, DE respectively The perpendicular from A onto EF has equation 1
(w – v)x – (e – f)y = a(w – v) – p(e – f).
(2.1)
The perpendicular from B onto FD has equation (u – w)x – (f – d)y = b(u – w) – q(f – d).
(2.2)
The perpendicular from C onto DE has equation (v – u)x – (d – e)y = c(v – u) – r(d – e).
(2.3)
3. The condition for orthologic triangles It is easy now to find the condition that the three lines in Section 2 are concurrent is a(v – w) + b(w – u) + c(u – v) + p(e – f) + q(f – d) + r(d – e) = 0.
(3.1)
Since this is the same equation when a, b, c are replaced by d, e, f and p, q, r are replaced by u, v, w it follows that if ABC is orthologic to DEF then DEF is orthologic to ABC. (There are too many variables for transitivity to be deduced from only one pair of such equations. Orthologic triangles date back to Steiner (1827).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
2
Article/CJB/2011/177
Special Pascal Lines Christopher Bradley Abstract: By considering the hexagon as two triangles in perspective and labelling their vertices as well as those of the defining conic, it is shown that two of the Pascal lines pass through points that are points of concurrence of three lines.
F
A
6 To F 5
R
1
L
E
Q
O B
To J
4
P 3
2
C
D
To J
N
K To M
H
M
Fig. 1 The two special Pascal lines LMN and HJK 1. Introduction The starting point in the construction of Fig.1 is to draw triangles 123 and 456 in perspective through a point O. The triangle ABC is formed by the sides 61 54 32 and the triangle DEF is formed from the sides 12 34 56. The six points A, B, C, D, E, F now lie on a conic (a proof of 1
which is given late in the article). We use homogeneous projective co-ordinates with 123 as triangle of reference and the vertex O of perspective as unit point O (1, 1, 1). We find six points L = BE^AF^24, M = CE^AD^26, N = BD^CF^46, H = AD^BF^35, J = BE^CD^15, K = AE^CF^13 and we prove LMN and HJK are Pascal lines, which we call special because they are points of concurrence of three lines. There are other points of concurrence of three lines which also define Pascal lines, and have a certain interest. 2. The co-ordinates of points and equations of sides and joins of triangles 123 and 456 With 123 of triangles of reference we have co-ordinates 1(1, 0, 0), 2(0, 1, 0), 3(0, 0, 1) and we take O, the vertex of perspective of the two triangles as O(1, 1, 1). It is the sufficiently general to take 4(1, a, a), 5(b, 1, b), 6(c, c, 1), where a, b, c are arbitrary constants. Equations of sides of the two triangles are now: 23: x = 0, 31: y = 0, 12: z = 0, 14: y = z, 25: z = x, 36: x = y, 34: y = ax, 61: y = cz, 35: x = by, 15: z = by, 24: z = ax, 26: x = cz, 45: a(b – 1)x + b(a – 1)y + (1 – ab)z = 0, 56: (bc – 1)x + b(1 – c)y + c(1 – b)z = 0, (2.1) 64: a(c – 1)x + (1 – ca)y + c(a – 1)z = 0. 3. The co-ordinates of points and equations of joins of triangles ABC and DEF The points A, B, C, D, E, F are now the intersections of lines obtained in Section 2 and their coordinates are A = 61^45: x = ab(c – 1) – bc + 1, y = ca(1 – b), z = a(1 – b). (3.1) B = 61 23: x = 0, y = c, z = 1. (3.2) C = 23^45: x = 0, y = ab – 1, z = b(a – 1). (3.3) D = 12^34: x = 1, y = a, z = 0. (3.4) E = 34^56: x = c(1 – b), y = ca(1 – b), z = ab(c – 1) – bc + 1. (3.5) F = 12^56: x = b(c – 1), y = bc – 1, z = 0. (3.6) The equations of the joins of these two triangles are AD: a(b – 1)x + (1 – b)y + (c – 1)(ab – 1)z = 0. (3.7) 2 2 2 2 AE: ca(a – 1)(b – 1)(bc – 1)x + (a b (c – 1) – a(b c(2c – 1) + 2b(1 – 2c) + c) + b2c2 – 2bc + 1)y + ca(b – 1)(c – 1)(ab – 1)z = 0. (3.8) AF: a(b – 1)(bc – 1)x + ab(1 – c)(b – 1)y + (ab(c – 1)(2bc – c – 1) – (bc – 1)2)z = 0. (3.9) BD: ax – y + cz = 0. (3.10) BE: (a – 1)(bc – 1) + (1 – b)(y – cz) = 0. (3.11) BF: (bc – 1)x + b(1 – c)(y – cz) = 0. (3.12) CD: ab(1 – a)x + b(a – 1)y + (1 – ab) z = 0. (3.13) 2
CE: CF:
(a2b(b(2c – 1) – c) + 2ab(1 – bc) + bc – 1)x + c(1 – b)(b(a – 1)y + (1 – ab))z = 0 (a – 1)(bc – 1)x + (1 – c)(b(a – 1)y + (1 – ab))z = 0.
(3.14) (3.15)
4. The points P, Q, R These points are each the concurrencies of three lines and their co-ordinates are as follows. P = AD^BE^25: x = b – 1, y = a(bc – 1) – c + 1, z = b – 1. (4.1) Q = BE^CF^14: x = (1 – c)(b – 1), y = (a – 1)(bc – 1), z = (a – 1)(bc – 1). (4.2) R = AD^CF^36: x = (1 – c)(ab – 1), y = (1 – c)(ab – 1), z = (a – 1)(b – 1). (4.3) Points P, Q, R do not lie on Pascal lines but have their properties in consequence of the perspective. 5. The conic ABCDEF From the co-ordinates of these points it is now possible to show they lie on a conic – this is, of course, a well known result. The conic has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (5.1) where we find u = 2a(1 – b)(1 – a)(1 – bc), v = 2b((1 – a)(1 – b)(1 – c), w = 2c((1 – b)(1 – c)(1 – ab), f = (1 – b)(1 – c)(ab(c + 1) – bc – 1), g = (1 – c)(a – 1)(ab(2bc – c – 1) – bc + 1), h = (1 – a)(1 – b)(ab(c – 1) + bc – 1).
(5.2) (5.3) (5.4) (5.5) (5.6) (5.7)
6. The Special Pascal lines LMN and HJK There are, of course, a vast number of Pascal lines, but there are only two that have the special property that the points defining them are points of concurrence of three lines in the figure (rather than two). We first give details of the line LMN. Their co-ordinates are L = AF^BE^24 : x = b – 1, y = a(2bc – c – 1) – bc + 1, z = a(b – 1). (6.1) M: AD^CE^26: x = c(b – 1), y = a(b(2c – 1) – c) – c + 1, z = b – 1. (6.2) N: BD^CF^46: x = 1 – c, y = a – c, z = a – 1. It may now be checked that L, M, N lie on the Pascal line with equation (a2(b(2c – 1) – c) + 2a(1 – bc) + bc – 1)x + (1 – b)(ca – 1)y + (a(b(2c2 – 2c + 1) – c2) – bc2 + 2c – 1)z = 0.
3
(6.3)
We now give details of the line HJK. Their co-ordinates are H = AD^BF^35: x = b(1 – c), y = 1 – c, z = b – 1. J = BE^CD^15 : x = 1 – b, y = a – 1, z = b( a – 1). K = AE^CF^13: x = (1 – c)(ab – 1), y = 0, z = (a – 1)(bc – 1).
(6.4) (6.5) (6.6)
It may now be checked that H, J, K lie on the Pascal line with equation (a – 1)(bc – 1)x – (ab2(c – 1) – b2c + 2b – 1)y + (1 – c)(1 – ab)z = 0.
(6.7)
It is also the case that P, R, M, H lie on AD, that P, Q, L, J lie on BE and Q, R, N, K lie on CE as may be checked from the definition of the six points L, M, N, H, J, K.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article: CJB/2011/178
More Circles centred on the Brocard Axis Christopher Bradley Abstract: Circles through B and C and the Brocard points lead to two circles centred on the Brocard axis and several concurrences. Z
Y
E
A U T
P
F
N
M
K
Q
' G B
R
S
C
X W
Fig.1 Circles centred on the Brocard axis Detailed description What is described now has been verified using areal co-ordinates with ABC as triangle of reference. We use the notation for the Brocard points Ω(1/b2, 1/c2, 1/a2) and Ω'(1/c2, 1/a2, 1/b2).
1
First circles BΩΩ' and CΩΩ' are drawn. BΩΩ' meets BC at R and AB at U. CΩΩ' meets BC at S and CA at T. It then transpires that points R, S, T, U are concyclic. The line TU is now drawn and it meets BΩΩ' at E, CΩΩ' at F and BC at G. It may now be proved that the four lines BE, RU, ST, CF are parallel. The diagonals of RSTU are now drawn and it turns out that US passes through Ω and the symmedian point K and that RT passes through Ω' and K. Points W, X, Y, Z are now obtained with W = BE^RT, X = ER^US^CT, Y = AB^TR^FS and Z = BE^US^AF. It then turns out that W, X. Y. Z are concyclic. Finally the four circles BΩΩ', CΩΩ', RSTU and WXYZ may be shown to have centres on the Brocard axis OK, where O is the circumcentre of ABC.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/179/2011
The Second Brocard Triangle Christopher Bradley
A
F O E
K ' D C
B
Fig. 1 Circle Intersections leading to the Second Brocard Triangle Abstract: Given the Brocard points Ω, Ω' in a triangle the point of concurrence of the intersection of the line AK with the 7-point circle and circles AΩC, BOC, AΩ'B (D in the above diagram) and two similar points E, F found by cyclic change of A, B, C forms what is known as the Second Brocard Triangle (David Monk – private communication). An analysis is the content of this document. 1. Intersections of AK, BK, CK with the 7-point triangle
1
The equation of the 7-point circle in areal co-ordinates is known to be b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0.
(1.1)
The line AK has equation c2y = b2z and this meets the 7-point circle at a second point D with coordinates (x, y, z), where x = b2 + c2 – a2, y = b2, z = c2. (1.2) Points E, F have co-ordinates which follow from those of D by cyclic change of a, b, c and x, y, z. 2. The circle BOC contains D Circles in areal co-ordinates have equations of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(2.1)
Circle BOC has an equation of this form with u = b2c2/(a2 – b2 – c2), v = 0, w = 0.
(2.2)
It may now be checked that point D lies on this circle. Similarly point E lies on circle COA and point F lies on circle AOB. 3. The circle AΩC contains D Circle AΩC has an equation of the form (2.1) with u = 0, v = – c2, w = 0.
(3.1)
It may now be checked that point D lies on this circle. Similarly point E lies on circle BΩA and point F lies on circle CΩB. 4. The circle AΩ'B contains D Circle AΩ'B has an equation of the form (2.1) with u = 0, v = 0, w = – b2.
(4.1)
It may now be checked that point D lies on this circle. Similarly point E lies on circle BΩ'C and point F lies on circle CΩ'A. The triangle DEF is called the Second Brocard Triangle (David Monk – private communication). Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
Article: CJB/180/2011
Circles through the Brocard points and the Circumcentre Christopher Bradley
Y
E
O
A
' F C
B
D X
Fig.1 Circles intersecting on the Circumcircle and Seven parallel lines Abstract: With Brocard points Ω and Ω', circles are drawn through AΩΩ', BOW', COW and they meet at a point D on the circumcircle. Points E, F are similarly defined. It is proved that the lines AD, BE, CD are all parallel to ΩΩ'. Further, if circle AΩΩ' meets AB at X and AC at Y then XY 1
is also parallel to ΩΩ', as are two further lines defined in a similar fashion to XY by cyclic change. 1. The circles AΩΩ', BOΩ', COW and their point of intersection D The equation of a circle in areal co-ordinates is of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(1.1)
The circle AΩΩ' has u = 0, v = c2a2(b2 – a2)/(a4 – b2c2), w = a2b2(c2 – a2)/(a4 – b2c2).
(1.2)
The circle BOΩ' has u = b2c2(c2 – b2)/(a4 – a2c2 + b4 – b2c2), v = 0, w = a2b2(c2 – a2)/(a4 – a2c2 + b4 – b2c2).
(1.3)
The circle COΩ has u = b2c2(b2 – c2)/(a4 – a2b2 + c4 – b2c2), v = c2a2(b2 – a2)/(a4 – a2b2 + c4 – b2c2), w = 0.
(1.4)
It may now be checked that these three circles have a common point D with co-ordinates (x, y, z), where x = a2(a2 – b2)(a2 – c2), y = b2(c2 – b2)(c2 – a2), z = c2(b2 – a2)(b2 – c2). (1.5) It may also be checked that D lies on the circumcircle of ABC with equation a2yz + b2zx + c2xy = 0.
(1.6)
In like manner circles BΩΩ', COΩ', AOΩ have a common point E which lies on the circumcircle of ABC and circles CΩΩ', AOΩ', BOΩ have a common point F which lies on the circumcircle of ABC. 2. The lines ΩΩ' and AD, BE, CF The equation of the line ΩΩ' is b2c2(a4 – b2c2)x + c2a2(b4 – c2a2)y + a2b2(c4 – a2b2)z = 0.
(2.1)
The equation of the line AD is c2(b2 – a2)(b2 – c2)y = b2(c2 – b2)(c2 – a2)z.
(2.2)
These lines meet on a point on the line at infinity with co-ordinates x = a2(b2 – c2), y = b2(c2 – a2), z = c2(a2 – b2).
(2.3)
It follows that these lines are parallel and from symmetry are also parallel to BE and CF. 2
3. Circle AΩΩ' meets AB at X and AC at Y and XY is also parallel to ΩΩ' The circle with equation derived from equation (1.2) meets AB at the point X with co-ordinates x = a2(a2 – b2), y = b2(a2 – c2), z = 0. (3.1) It meets AC at the point Y with co-ordinates x = a2(a2 – c2), y = 0, z = c2(a2 – b2).
(3.2)
The equation of XY may now be obtained and is b2c2(a2 – b2(a2 – c2)x = c2a2(a2 – b2)2 + a2b2(c2 – a2)2.
(3.3)
It may now be shown that XY is also parallel to ΩΩ' as are two further lines joining the intersections of circles BΩΩ' and CΩΩ' with appropriate sides of ABC. This makes 7 parallel lines in all. See Fig.1. Of course, these lines all being perpendicular to the Brocard axis are all parallel to the tangents at O and K to the 7-point circle.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/181
Circles through the Brocard points and the Symmedian Point Christopher Bradley
E
F
A
K ' D
B
C
Fig. 1 Circles through the Brocard points and Symmedian point provide a line parallel to ΩΩ' 1
Abstract: Circles BΩK, CΩ'K both pass through the same point D on BC. E and F are defined by cyclic change and lie on CA and AB. DEF turn out to lie on a line parallel to the tangent at K to the 7-point circle (and hence perpendicular to the Brocard axis) 1. Circles BΩK and CΩ'K and the point D Circles in areals have an equation of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0 For circle BΩK we find u = b2c2(c2 + a2 – 2b2)/{(a2 + b2 + c2)(b2 – c2)}, v = 0, w = a2b2(a2 + b2 – 2c2)/{(a2 + b2 + c2)(c2 – b2)}. For the circle CΩ'K we find u = b2c2(b2 + a2 – 2c2)/{(a2 + b2 + c2)(c2 – b2)}, v = a2c2(a2 + c2 – 2b2)/{(a2 + b2 + c2)(b2 – c2)}. w = 0. These two circles intersect at K and again at a point D on BC with co-ordinates x = 0, y = – b2(a2 + b2 – 2c2), z = c2(c2 + a2 – 2b2).
(1.1)
(1.2)
(1.3)
(1.4)
E and F lie on CA and AB respectively and have co-ordinates obtained from (1.4) by cyclic change of x, y, z and a, b, c.. 2. The line DEF is parallel to the tangent at K to the Brocard circle The equation of the line DEF is b2c2(b2 + c2 – 2a2)x + c2a2(c2 + a2 – 2b2)y + a2b2(a2 + b2 – 2c2)z = 0.
(2.1)
The equation of the line ΩΩ' is b2c2(a4 – b2c2)x + c2a2(b4 – c2a2)y + a2b2(c4 – a2b2)z = 0.
(2.2)
These two lines meet at a point on the line at infinity with co-ordinates x = a2(b2 – c2), y = b2(c2 – a2), z = c2(a2 – b2).
(2.3)
It follows that line DEF and the tangent at K to the Brocard circle are parallel.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
Article: CJB/2011/182
Problems requiring Proof Christopher Bradley
A T
V
W S
B
U
R
C
Fig. 1 Problem 1 Let ABC be a triangle and let Σ be a conic not passing through A, B or C. The tangents to Σ from A meet BC at R and U, from B meet CA at S and V, and from C meet AB at T and W. Prove that a conic passes through R, S, T, U, V, W.
1
42 23 13
12
T
B 21
A Q 14
R
41 P 31
34
U D V C
32
W
S 43 24
Fig. 2 Problem 2 ABCD and PQRS are two quadrilaterals which are in axial perspective, that is T = AB^PQ, U = BC^QR, V = CD^RS, W = DA^SP are collinear. In the above figure the other twelve intersections of the sides of the quadrilaterals are shown with notation exemplified by 13 = AB^RS, 42 = DA^QR etc. It is then the case that four conics passing through six points each, two conics passing through each of these points exist. Thus 23, 24, 32, 34, 42, 43 lie on conic Σ1, 13, 14, 31, 34, 42, 43 lie on conic Σ2, 12, 14, 21, 24, 41, 42 lie on conic Σ3 and 12, 13, 21, 23, 31, 32 lie on conic Σ4. (The two quadrilaterals do not need to be convex.)
2
4
D S
A
13 F 23
24 14 12
34 43 41
21 B
2
1
32 T
31
P
C Q
42 R 3
Fig. 3 Problem 3 This is similar to Problem 2, but this time the Quadrangles ABCD and PQRS are such that with F a Cevian point of ABCD and T a Cevian point of PQRS then the four intersections AF^PT, BF^QT, CF^RT and DF^ST lie on a line. Then the other twelve points of intersection of pairs of Cevian lines have the property that four conics exist, two through each of the twelve points and six points of intersection on each conic.
3
A S
D W
V
O P
R
T
U C
Q
B
Problem 4 ABCD is a cyclic quadrilateral centre O and P, Q, R, S are the midpoints of the sides AB, BC, CD, DA. Prove that a circle centre O can be found with points T, U, V, W lying on it such that PTUQ, QUVR, RVWS, SWTP are all circles
4
Article: CJB/2011/183
Conics and Triangles in Perspective Christopher Bradley
A Q W
Y
R D
Z
K E V
F B
U
X
C
P
Fig. 1 Abstract: It is shown how a pair of triangles in perspective lead to a conic and conversely. 1. Introduction Given a triangle ABC and a point K, not on the sides, choose points D, E, F on the Cevian lines AK, BK, CK respectively. Let DE meet BC in U and CA in Y. Let EF meet AB in Z and CA in V and let FD meet BC in X and AB in W. Then we prove that a conic passes through the six points U, V, W, X, Y, Z. Furthermore ZU and VX meet at a point P on AK, with Q, R defined similarly so that conic cuts the sides of all three triangles in six points. Conversely if we strat with a triangle ABC and a conic cutting each side in two points, a triangle DEF arises and AD, BE, CF meet in a point K so that triangles ABC and DEF are in perspective. Finally Cantor’s relationship is established: (BU/UC)(BX/XC)(CV/VA)(CY/YA)(AW/WB)(AZ/ZB) = 1 (1.1)
is verified. (During the course of the analytic proof it is discovered how to write down the equation of a conic which meets the sides of a triangle in points with rational co-ordinates.) 2. The ten points We use areal co-ordinates with ABC the triangle of reference and we take K to have co-ordinates (f, g, h). Since D lies on AK we take it to have co-ordinates D(u, g, h) and similarly E(f, v, h) and F(f, g, w). The equation of EF is then (vw – gh)x + f(h – w)y + f(g – v)z = 0.
(2.1)
We can now write down the co-ordinates of Z= EF^AB which are Z(f(h – w), gh – vw, 0) and V = CA^EF which are V(f(g – v), 0, gh – vw). The co-ordinates of the other four points may now be written down using cyclic change of x, y, z and f, g, h and u, v, w. They are X(0, g(f – u), hf – wu), W(hf – wu, g(h – w), 0), Y(fg – uv, 0, h(g – v)), and U(0, fg – uv, h(f – u)). 3. The conic through U, V, W, X, Y, Z It may now be shown that a conic passes through the six points U, V, W, X, Y, Z with an equation of the form lx2 + my2 + nz2 + 2pyz + 2qzx + 2rxy = 0 (3.1) where l = 2gh(f – u)(g – v)(h – w)(vw – gh), m = 2hf(f – u)((g – v)(h – w)(wu – hf), n = 2fg(f – u)(g – v)(h – w)(uv – fg), p = f(g – v)(h – w)(2f2gh – 2ufgh – hfuv + u2(gh + vw)), q = g((h – w)(f – u)(2g2hf – 2vfgh – fgvw + v2(hf + wu)), r = h(f – u)(g – v)(2h2fg – 2wfgh – ghwu + w2(fg + uv)).
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
4. The triangle PQR The line ZU has equation h(f – u)(vw – gh)x + hf(h – w)(f – u)y + f(h – w)(uv – fg)z = 0.
(4.1)
The equation of XV is g(f – u)(gh – vw)x + f(g – v)(hf – wu)y + fg(g – v)(u – f)z = 0.
(4.2)
These lines intersect in the point P with co-ordinates (x, y, z), where x = fu(g – v)(w – h)/{(f – u)(gh – vw)}, y = g, z = h.
(4.3)
P clearly lies on AK with equation hy = gz. Points Q, R have co-ordinates that may be written down from (4.3) by cyclic change of x, y, z and f, g, h and u, v, w. Also Q lies on BK and R on CK. 5. Cantor’s equation From the co-ordinates of U, V, W, X, Y, Z we have BU/UC = h(f – u)/(fg – uv), BX/XC = (hf – wu)/(g(f – u)), CV/VA = f(g – v)/(gh – vw), CY/YA = (fg – uv)/{h(g – v)}, AW/WB = g(h – w)/(hf – wu), AZ/ZB = (gh – vw)/{f(h – w)}. The product of all these fractions leads to equation (1.1) as required. 6. The converse The converse, that if one begins with triangle ABC and a conic cutting all three sides in pairs of points from which triangles DEF and PQR are in perspective with each other and ABC follows, since the equation of the conic in Section 3 is general (having five independent ratios). If any pairs coincide the conic has tangency and conversely.
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Article CJB/2011/184
Cevian Perspectivity Christopher Bradley A
33 R
21 13 23 D X S 32 31 12
Q
22 B
C
P
11
Fig. 1 Abstract: Triangle ABC and PQR with Cevian points D and S respectively are said to be in Cevian perspective if the points 11 = AD^PS, 22 = BD^QS, 33 = CD^RS are collinear. If points are labelled as follows: 12 = AD^QS,13 = AD^RS, 21 = BD^PS, 23 = BD^RS, 31= CD^PS, 32 = CD^QS, then the following results hold: (i) triangles 13 23 21 and 31 32 12 are in (ordinary) perspective through a vertex X, (ii) the points 13 23 21 31 32 12 are co-conic and (iii) X lies on the line SD. 1. Introduction Projective co-ordinates are used with ABC the triangle of reference and S the unit point. Whilst (i) and (ii) are easily established by pure methods (iii) is the interesting feature of Cevian 1
perspectivity and for this an analytic proof is required. For ease of working we suppose that P, Q, R lie on BC, CA, AB respectively, but the theorems are true under more general conditions, as CABRI II plus confirms. 2. The main points and lines We suppose D has the co-ordinates D(a, b, c) (not the sides of a triangle). Also let P(0, u, 1 – u), Q(1 – v, 0, v), R(w, 1 – w, 0) and S(1, 1, 1). Lines AD, BD, CD have equations bz = cy, cx = az, ay = bx respectively. Lines PS, QS, RS have equations as follows: PS: (1 – 2u)x + (u – 1)y + uz = 0, (2.1) QS: vx + (1 – 2v)y + (v – 1)z = 0, (2.2) RS: (w – 1)x + wy + (1 – 2w)z = 0. (2.3) The co-ordinates of the points 11 = AD^PS 22 = BD^QS, 33 = CD^RS may now be obtained and are: 11: x = uc + (u – 1)b, y = (2u – 1)b, z = (2u – 1)c, (2.4) 22: x = (2v – 1)a, y = va + (v – 1)c, z = (2v – 1)c, (2.5) 33: x = (2w – 1)a, y = (2w – 1)b, z = wb + (w – 1)a. (2.6) The condition for 11 22 33 to be collinear, so that a Cevian perspectivity exists, is given by the equation w = (1/k)(a2(b(u(3v – 2) – v + 1) + cv(u – 1)) – ac(b(u(9v – 5) – 5v + 3) + c(1 – v)(u – 1)) + bc(2v – 1)(b(u – 1) + cu)), (2.7) where k = a2(b(u(3v – 2) – v + 1) + cv(3u – 2)) + a(b2(u(3v – 2) – v + 1) – bc(9u(2v – 1) – 9v + 5) + c2(v – 1)(3u – 2)) + bc(3v – 1)(b(u – 1) + cu). (2.8) It is now possible to obtain the co-ordinates of the six main points, which are 12 = AD^QS: x = b(2v – 1)) + c(1 – v), y = bv, z = cv, 23 = BD^RS: x = aw, y = a(1 – w) + c(2w – 1), z = cw, 31 = CD^PS: x = au, y = bu, z = b(1 – u) + a(2u – 1), 21 = BD^PS: x = a(u – 1), y = a(2u – 1) – cu, z = c(u – 1), 32 = CD^QS: x = a(v -0 1), y = b(v – 1), z = b(2v – 1) – av, 13 = AD^RS: x = c(2w – 1) – bw, y = b(w – 1), z = c(w – 1).
(2.9) (2.10) (2.11) (2.12) (2.13) (2.14)
3. The conic and the condition it passes through all six main points The equation of a conic is px2 + qy2 + rz2 + 2fyz + 2gzx + 2hxy = 0, 2
(3.1)
and the values of the constants p, q, r, f, g, h so that it passes through 12, 23, 31, 21, 32 are p = 2cv(a3v(w – 1)(2u – 1) + a2(1 – 2u)(2b(w – 1)(2v – 1) + c(vw + w – 1)) – a(b2(u(v(w + 2) – w – 1) + v(w – 2) + 1) – bc(u(v(19w – 10) – 9w + 5) + 2(1 – 2w)(2v – 1)) + c2(v – 1)(2w – 1)(2u – 1)) – bc(b(u – 1) + cu)(v(3w – 2) – w + 1)), (3.2) 2 q = 2acw(1 – u)(av + b(1 – 2v) + c(v – 1)) , (3.3) r = 2acu(1 – v)(av + b(1 – 2v) + c(v – 1))(a(w – 1) + bw + c(1 – 2w)), (3.4) 2 2 f = – a(a b(u(3v – 2) – v + 1)(v(3w – 2) – w + 1) – c(u(v (7w – 5) + 2v(3 – 4w) + 2(w – 1)) + (1 – v)(v(2w – 1) – w + 1))) + ac(c(u(2v2(5w – 3) + v(8 – 11w) + 3(w – 1)) + 2(1 – w)(v – 1))) + c(b2w(2v – 1)(uv + v – 1) + bc(v – 1)(u(v(w – 2) – w + 1) + v(2 – 5w) + 3w – 1) – c2(v – 1)2(u(w – 1) – 2w + 1))), (3.5) 3 2 2 g = a cv(w – 1)(u(v – 2) – v + 1) + a (b (u(3v – 2) – v + 1)(v(3w – 2) – w + 1) – 2 bc(u(v (7w – 6) + 11v(1 – w) + 4(w – 1)) + 2(1 – w)(2v2 – 3v + 1)) – c2(u(v – 2) – v + 1)(vw + w – 1)) – ac(b2(u(v2(17w – 10) + v(11 – 19w) + 5w – 3) + v2(2 – 5w) + v(7w – 3) – 2w + 1) + bc(1 – 2v)(u(v(7w – 4) – 9w + 5) + 2(1 – 2w)(v – 1)) + c2(v – 1)(2w – 1)(u(v – 2) – v + 1)) + bc2u(b(2v – 1) + c(1 – v))(v(3w – 2) – w + 1), (3.6) 3 2 2 2 h = c(a v (u(w + 1) – 1) + a (b(u(5v – 2) – 3v + 1)(v(w – 2) – w + 1) – c(2u(v (2w – 3) + v(4 – 3w) + w – 1) + 2v2 + v(3w – 4) – w + 1)) + a(b2(4v2 – 4v + 1)(u(w + 1) – 1) + bc(1 – 2v)(u(v(11w – 4) – 5w + 1) – 2(v(3w – 1) – w)) + c2(u(v2(13w – 7) + v(8 – 15w) + 2(2w – 1)) + (1 – 2w)(3v2 – 4v + 1))) + bc(u – 1)(b(2v – 1) + c(1 – v))(v(3w – 2) – w + 1)). (3.7) The condition (2.7) now ensures that this conic also passes through the point 13. It is worth mentioning the fact that three general lines intersect three other general lines in nine points and three of the nine points such as 11, 22, 33 are collinear, then except in special circumstances, the other six points lie on a conic (or two other straight lines). This is a general theorem in quantitative geometry (see, for example, Sylvester’s Geometry Ancient and Modern). 4. The point X The equation of the line SD is (b – c)x + (c – a)y + (a – b)z = 0.
(4.1)
In view of the line 11 22 33 (regarded as a Desargues’ axis it follows that triangles 13 23 21 and 31 32 12 are in perspective with a vertex X and it is found that the co-ordinates of X are x = (a2(b(u(3v – 2) – v + 1) – c(u(v – 2) + 1)) + 2ac(1 – 2u)(b(2v – 1) + c(1 – v)) + c(b2(u – 1)(2v – 1) + bc(uv + v – 1) + c2u(1 – v))), (4.2) 2 2 2 y = a cv(2u – 1) + a(b (u(3v – 2) – v + 1) + 2bc(1 – 2v)(2u – 1) + c (u(v – 2) – v + 1)) – 2 c(b (u(v – 1) + v) + 2bcu(1 – 2v) + c2u(v – 1)), (4.3) 2 2 2 z = c(a v(2u – 1) – 2av(b(u – 1) + cu) + b (u – 1)(2v – 1) + 2bc(1 – v)(u – 1) + c (u(2v – 1) – v + 1)). (4.4) 3
It is now straightforward to show that X lies on SD, which is a slightly surprising result. The more general situation is shown in Fig. 2 below.
A
R
22
13 Q 12
X
S 21
32
D
23
31
B
C
33
P
11
Fig. 2 Another point of interest is that as a result of the perspective of the two triangles that line 32 13 intersects 23 31 at a point on line 11 22 33 as do lines 12 31 and 13 21.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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1 Article: CJB/2011/185
The Four Conic Theorem Christopher Bradley
11
12
21
31
41
14
13
22
23
32
42
24
33
34
43
44
Fig. 1 The Four Conics on a Square Grid Abstract: Given two sets of four lines which intersect one on one in four points in a straight line, then the twelve remaining points carry four conics in which through point two conics pass. 1. Introduction
2 We provide an account of the four conic theorem when the two sets of four lines form a square grid. Then the sixteen points may be labelled (x, y) for x = 1 to 4 and y = 1 to 4. We suppose further that the points of intersection 11 22 33 44 form the line. The theorem now is that the following sets of six points are each co-conic: (i) 23 24 34 43 42 32 (ii) 13 14 34 43 41 31 (iii) 12 14 24 42 41 21 and (iv) 12 13 23 32 31 21. We use Cartesian co-ordinates and provide the equations of the four conics. 2. The Four Conics Conic (i) has equation 3(x + y – 6)2 + (x – y)2 = 4.
(2.1)
x2 + y2 + xy – 8x – 8y + 19 = 0
(2.2)
x2 + y2 + xy – 7x – 7y + 14 = 0.
(2.3)
3(x + y – 4)2 + (x – y)2 = 4.
(2.4)
Conic (ii) has equation Conic (iii) has equation Conic (iv) has equation
The four conic theorem more generally holds when the lines are distinct and intersect one and one in four collinear points. Then a point such as 23 means the intersection of line 2 of the first set of lines with line 3 of the second set of lines. The theorem generalizes. For example in the 5 x 5 case there are 10 conics passing through 20 points with 3 conics passing through each point and in the 6 x 6 case there are 20 conics passing through 30 points with 4 conics passing through each point and in the 7 x 7 case there are 35 conics passing through 42 points with 5 conics through each point.
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Article: CJB/2011/186
6 Points and 4 perspectives Christopher Bradley Abstract: If 6 points provide 1 pair of triangles in perspective then there are 3 more pairs of perspective triangles, with the same vertex, in the same diagram. Their Desargues’ axes create the sides of a triangle and a related transversal. (It is assumed the original pair of triangles are not in triple perspective.)
A R cy F
Q 13
21 23
cx
by Y
32 12 B
E
O
X
ax
bx
31 P D
ay
C
Fig. 1 The four perspectives and their Desargues’ axes BCP, CQA, ARB and PQR 1. Introduction The implication of the abstract is that we start from two triangles 13 23 21 and 31 32 12 in perspective with vertex of perspective O. However the triangle ABC and the two Cevian points X and Y then emerge. It is easier from a mathematical point of view to proceed from triangle ABC and two Cevian points X and Y and to derive the perspectives. This is because theorems and their converses are both true. (It is the case, however, that if two triangles are in triple perspective, then triangle ABC degenerates into a line and P, Q, R lie on that line, which is a whole new story involving the Brocard porism.)
1
2. ABC, X, Y and the six points 13, 23, 21, 31, 32, 12 We use areal co-ordinates throughout with ABC as triangle of reference and X, Y having coordinates X(a, b, c) and Y(f, g, h). Here it must be emphasized that a, b, c are not the sides of the triangle ABC, but arbitrary constants, just as f, g, h are also. We also remark, in case it has not been understood, that if UVW and U'V'W' are in perspective through a vertex O then so are triangles UV'W and U'VW' and so are triangles UVW' and U'V'W and so are triangles UV'W' and U'VW, making altogether 4 perspectives and hence though the vertex is the same in all four cases, the four Desargues’ axes are distinct (though only 6 points occur). The equations of the Cevians are: AX: bz = cy, BX: az = cx, CX: ay = bx, AY: gz = hy, BY: fz = hx, CY: fy = gx.
(2.1)
We can now define the six points and list their co-ordinates. 13 = AX^CY: x = bf, y = bg, z = cg. 12 = AX^BY: x = cf, y = bh, z = ch. 21 = BX^AY: x = ah, y = cg, z = ch. 23 = BX^CY: x = af, y = ag, z = cf. 31 = CX^AY: x = ag, y = bg, z = bh. 32 = CX^BY: x = af, y = bf, z = ah.
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7)
3. The vertex O and the points P, Q, R It is clear from Figure 1 that triangles 23 21 13 and 32 12 31 are in perspective with vertex O. To find the co-ordinates of O we determine the equations of lines 12 21 and 13 31 which are respectively ch(bh – cg)x + ch(ah – cf)y + (c2fg – abh2)z = 0, (3.1) and bg(bh – cg)x + (acg2 – b2fh)y + bg(bf – ag)z = 0.
(3.2)
These lines meet at the point O with co-ordinates O (af(bh + cg), bg(ah + cf), ch(ag + bf)). (3.3) It may now be checked that the line 23 32 with equation (a2gh – bcf2)x + af(cf – ah)y + af(bf – ag)z = 0 also passes through O.
2
(3.3)
The line 13 21 has equation cg(cg – bh)x + ch(bf – ag)y + bg(ah – cf)z = 0.
(3.4)
This meets BC, x = 0, at P with co-ordinates P(0, bg(ah – cf), ch(ag – bf)).
(3.5)
It may now be checked that P also lies on the line 31 12 with equation bh(cg – bh)x + ch(bf – ag)y + bg(ah – cf)z = 0.
(3.6)
Similarly the point Q = 21 32^ 23^12 ^ CA has co-ordinates Q(af(cg – bh), 0, ch(ag – bf)) (3.7) and the point R = 23 12 ^31 23^AB has co-ordinates R(af(bh – cg), bg(ah – cf), 0).
(3.8)
4. The four Desargues’ axes of perspective It may now be seen from Fig. 1 and Section 3 that triangles 13 23 21 and 31 32 12 are in perspective with axis BCP. Also that triangles 13 32 21 and 31 23 12 are in perspective with axis PQR. Also that triangles 13 23 12 and 31 32 21 are in perspective with axis CQA. Also that triangles 13 32 12 and 31 23 21 are in perspective with axis ARB. 5. Further results The foot of the Cevian AX we label ax, with bx, cx, ay, by, cz similarly defined. It is well known that these six points lie on a conic. Its equation is bcghx2 + cahfy2 + abfgz2 – af(bh + cg)yz – bg(ah + cf)y – ch(ag + bf)xy = 0. (5.1) It may also be shown that if X and Y are the centroid and orthocentre of ABC then the vertex O is the symmedian point of ABC and similarly if X and Y are the centroid and Gergonne’s point then O is the incentre of ABC. Finally the transversal PQR is such that if the harmonic conjugate of P with respect at ax, ay is P', with Q', R' similarly defined then P', Q', R' are the feet of the Cevians of the vertex O as Cevian point. That is P' = AO^BC etc. Proof is left to the reader.
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Article: CJB/2011/187
A Perspective in a pair of Cyclic Quadrilaterals Christopher Bradley Abstract: When two cyclic quadrilaterals in the same circle are in perspective then an axis of perspective joining the intersections of all six pairs of corresponding sides exists.
ad P ac
ab A'
B'
C'
A
D' cd bd D
O
bc C B
Fig. 1 1. Introduction We use Cartesian co-ordinates with circle having equation x2 + y2 = 1.
(1.1)
The point P is the vertex of perspective and has co-ordinates P(u, 0), u ≠ 1. The point A is given the co-ordinates x = (1 – a2)/(1 + a2), y = 2a/(1 + a2). (1.2) Points B, C, D have similar co-ordinates, but with parameters b, c, d respectively. 2. The second cyclic quadrilateral The line PA has equation 2ax + ((u + 1)a2 + u – 1))y = 2au. 1
(2.1)
PA meets the circle again at a point A' with co-ordinates (x, y), where x = {(u + 1)2a2 – (u – 1)2}/{(u + 1)2a2 + (u – 1)2}, y = 2a(u2 – 1)/{(u + 1)2a2 + (u – 1)2}.
(2.2) (2.3)
Points B', C', D' have similar co-ordinates with parameters b, c, d. 3. The axis of perspective The equation of AB is (1 – ab)x + (a + b)y = 1 + ab.
(3.1)
The equation of A'B' is ((u + 1)2ab – (u – 1)2)x + (u2 – 1)(a + b)y = (u + 1)2ab – (u – 1)2.
(3.2)
These meet at the point ab with co-ordinates x = 1/u, y = {(u + 1)ab + (u – 1)}/{(u(a + b)}.
(3.3)
The other points ac, ad, bc, bd, cd also lie on the line with equation x = 1/u. This axis of perspective is, of course, the polar of P with respect to the circle.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/188
Are three Triangles ever in Mutual Perspective Christopher Bradley Abstract: The question in the title has the answer ‘Yes’, for example three triangles in the Brocard porism are in fact in triple reverse perspective with each other. However, we study in this very short article the condition that three triangles must obey if they are to be in mutual perspective. (Perspectives, of course, are symmetric, but not generally associative.)
A
J W
B C E
P
F D T V
U
Fig. 1 Three triangles in mutual perspective 1. The first perspective We use projective co-ordinates with ABC the triangle of reference and the vertex of the first perspective at the unit point P. Triangle ABC is in perspective with triangle DEF with vertex P. The co-ordinates of D, E, F are therefore D(p, 1, 1), E(1, q, 1), F(1, 1, r) (1.1) for some constants p, q, r. 2. The second perspective 1
We suppose next that triangle ABC is in perspective with another triangle UVW with vertex of perspective J with co-ordinates J(u, v, w). The co-ordinates of U, V, W may therefore be taken as U(d, v, w), E(u, e, w), F(u, v, f), (2.1) where d, e, f are constants. What transpires is that if d, e, f are arbitrary the in general triangle DEF and UVW are not in perspective. However, as we shall shortly see, if D is moved along AJ there is one position only for D which produces a perspective (vertex T in Fig.1). 3. The third perspective We find the condition that DU, EV, FW are concurrent. The equations of these three lines are DU: (w – v)x + (d – pw)y + (pv – d)z = 0, (3.1) EV: (qw – e)x + (u – w)y + (e – qu)z = 0, (3.2) FW: (f – rv)x + (ru – f)y + (v – u)z = 0. (3.3) The condition that these three lines are concurrent, which may be regarded as a linear equation for d, in terms of p, q, r, u, v, w, e, f is d(e(r – 1)(u – v) + f(q – 1)(w – u) + q(ru(v – w) + w(u – v)) + rv(w – u)) + e(f(p – 1)(v – w) + p(w(u – v) – rv(u – w)) + ru(v – w)) + f(p(qw(u – v) – v(u – w)) + qu(v – w)) – p(qw2(u – v) + rv2(w – u)) + (w – v)(qru2 + (u – v)(w – u)) = 0. (3.4) In Fig. 1 the Desargues’ axes of perspective are drawn. No particular additional properties emerge when the perspectives are mutual, but none was expected.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/189
The Midpoint Conic Christopher Bradley Abstract: It is familiar that the feet of a pair of Cevians are co-conic. We prove here that their midpoints are also co-conic.
A
W I
G
X
H
Z E J
K
T
V
Y B
F
U
C
Fig. 1 The midpoints of the Cevians lie on a conic 1. The Cevians We take the Cevian points E and T to have co-ordinates E(f, g, h) and T(u, v, w). The feet of the Cevians through E therefore have co-ordinates F(0, g/(g + h), h/(g + h)), G(f/(h + f), 0, h/(h + f)) and H(f/(f + g), g/(f + g), 0). And the feet of the Cevians through T have co-ordinates U(0, v/(v + w), w/(v + w)), V(u/(w + u), 0, w/(w + u)), W(u/(u + v), v/(u + v), 0). 2. The conic FGHUVW This conic has an equation of the form ax2 + by2 + cz2 + 2pyz + 2qzx + 2rxy = 0, 1
(2.1)
where we find by substitution of co-ordinates that a = 2ghvw, b = 2hfwu, c = 2fguv, p = – fu(gw + hv), q = – gv(hu + fw), r = hw(fv + gu). (2.2) 3. The midpoints IJKXYZ of the Cevians The co-ordinates of these points are as follows: I, the midpoint of AF: x = ½, y = g/{2(g + h)}, z = h/{2(g + h)}, J, the midpoint of BG: x = f/{2(h + f)}, y = ½, z = h/{2(h + f)}, K, the midpoint of CH: x = f/{2(f + g)}, y = g/{2(f + g)}, z = ½, X, the midpoint of AU: x = ½, y = v/{2(v + w)}, z = w/{2(v + w)}, Y, the midpoint of BV: x = u/{2(w + u)}, y = ½, z = w/{2(w + u)}, Z, the midpoint of CW: x = u/{2(u + v)}, y = v/{2(v + u)}, z = ½.
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
4. The conic IJKXYZ This has an equation of the form (2.1), where we find a = f(u + v + w) + g(u + v – w) + h(u – v + w), b = g(u + v + w) + h(v + w – u) + f(v – w + u), c = h(u + v + w) + f(w + u – v) + g(w – u + v), p = fu – (g + h)(v + w), q = gv – (h + f)(w + u), r = hw – (f + g)(u + v).
(4.1) (4.2) (4.3) (4.4) (4.5) (4.6)
There is no conic through the quarter way down points of the Cevians and there appears to be no obvious geometrical link between the conics in Sections 2 and 4 (such as a relationship of their centres and the Cevian points). It should also be noted that a midpoint conic also exists when the six points FGHUVW are the intersections of any conic with the sides of a triangle. The proof of which is left to the reader.
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3
Article: CJB/2011/190
The Brocard Circles’ Twin Conic Christopher Bradley Abstract: The construction of the Triplicate Ratio Circle and the Brocard Circle is replicated but starting with the circumcentre O rather than the symmedian point K. It is found that the Brocard Circle is replaced by a conic (the twin conic) passing through O and K and having centre the midpoint of OK. This result leads to the concept of a conjugation whose properties we briefly outline.
A
E'
F W' N F'
O
X
L K
E
W M
B
D
D'
C
Fig. 1 Twin conics of the Triplicate Ratio Circle and the Brocard circle 1. Introduction
1
The twin conics are defined as follows: Lines parallel to the sides of a triangle ABC are drawn though its circumcentre O (rather than K as in the case of the Triplicate Ratio Circle and the Brocard Circle) meeting the side BC in D and D', meeting the side CA in E and E' and meeting the side AB in F and F', so that F'E is parallel to BC, D'F is parallel to CA and E'D is parallel to AB. See Fig. 1. We prove that the six points D, E, F, D', E', F' lie on a conic (the twin of the Triplicate Ratio Circle). Its centre is the midpoint of OK, which is also the centre of the Triplicate Ratio Circle. Lines through A, B, C parallel to FD, DE, EF respectively meet at W and lines through A, B, C parallel to D'E', E'F', F'D' respectively meet at W' (In the construction of the Brocard circle W and W' correspond to the Brocard points). The points L, M, N are defined as follows: L = BW^CW', M = CW^AW', N = AW^BW'. We then show that the conic through W, W', L, M, N passes through O and K and also has centre the midpoint of OK, which is also the centre of the Brocard circle. Clearly the theory of the Brocard circle and its twin is related in a way that is not a coincidence. Further investigations show that a conjugation is involved and we briefly outline its properties. Isogonal conjugation relates points (x, y, z) with (a2/x, b2/y, c2/z) and isotomic conjugation relates (x, y, z) with (1/x, 1/y, 1/z). This particular conjugation relates (x, y, z) with (x(y + z – x). y(z + x – y), z(x + y – z)) and it will immediately be seen that O and K are conjugates. The incentre I(a, b, c) and the Mittenpunkt Mi(a(b + c – a), b(c + a – b), c(a + b – c)) are also conjugates. 2. Analysis of the Twin Conic of the Triplicate Ratio Circle The line through O parallel to BC has an equation of the form x + s(x + y + z) = 0,
(2.1)
where s must be chosen so that the line passes through O. The appropriate value of s is s = a2(b2 + c2 – a2)/(a4 + b4 + c4 – 2b2c2 – 2c2a2 – 2a2b2).
(2.2)
Thus the equation of the line is (a2(b2 + c2) – (b2 – c2)2)x – a2(b2 + c2 – a2)(y + z) = 0.
(2.3)
This line meets CA at E with co-ordinates x = a2(b2 + c2 – a2), y = 0, z = a2(b2 + c2) – (b2 – c2)2.
(2.4)
And it meets AB at F' with co-ordinates x = a2(b2 + c2 – a2), y = a2(b2 + c2) – (b2 – c2)2. z = 0.
(2.5)
2
Similar analysis provides co-ordinates of D' on BC, which are x = 0, y = b2(c2 + a2 – b2), z = b2(c2 + a2) – (c2 – a2)2.
(2.6)
And the co-ordinates of F on AB are x = b2(c2 + a2) – (c2 – a2)2, y = b2(c2 + a2 – b2).
(2.7)
Similarly the co-ordinates of D on BC are x = 0, y = c2(a2 + b2) – (a2 – b2)2, z = c2(a2 + b2 – c2).
(2.8)
And the co-ordinates of E' on CA are x = c2(a2 + b2) – (a2 – b2)2, y = 0, z = c2(a2 + b2 – c2).
(2.9)
3. The conic through D, E, F, D', E', F' and its centre This has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where we find u = 2b2c2(c2 + a2 – b2)(a2 + b2 – c2)(a2(b2 + c2) – (b2 – c2)2), v = 2c2a2(a2 + b2 – c2)(b2 + c2 – a2)(b2(c2 + a2) – (c2 – a2)2, w = 2a2b2(b2 + c2 – a2)(c2 + a2 – b2)(c2(a2 + b2) – (a2 – b2)2, f = – a2(b2 + c2 – a2)(a8 – 3a6(b2 + c2) + a4(3b4 + 4b2c2 + 3c4) – a2(b – c)2(b2 – c2)2), g = – b2(c2 + a2 – b2)(b8 – 3b6(c2 + a2) + b4(3c4 + 4c2a2 + 3a4) – b2(c – a)2(c2 – a2)2), h = – c2(a2 + b2 – c2)(c8 – 3c6(a2 + b2) + c4(3a4 + 4a2b2 + 3b4) – c2(a – b)2(a2 – b2)2.
(3.2) (3.3) (3.4) (3.5) (3.6) (3.7)
In terms of u, v, w, f, g, h the centre of the conic (2.1) has co-ordinates (x, y, z), where x = vw – gv – hw – f2 + fg + hf, y = wu – hw – fu – g2 + gh + fg, z = uv – fu – gv – h2 + hf + gh.
(3.8) (3.9) (3.10)
After some massive cancellation we find the centre of the above conic has co-ordinates (x, y, z), where x = a2(b2(2c2 + a2) + a2(c2 – a2)), y = b2(c2(2a2 + b2) + b2(a2 – b2)), z = c2(a2(2b2 + c2) + c2(b2 – c2)).
(3.11) (3.12) (3.13)
3
It may now be checked that this is the point X, the midpoint of OK. 4. The analysis of the Twin Conic of the Brocard circle
The equation of DE is (a4 – a2(2b2 + c2) + b2(b2 – c2))(a2(b2 + c2) – (b2 – c2)2)x – a2(b2 + c2 – a2)(c2(a2 + b2 – c2)y + (a4 – a2(2b2 + c2) + b2(b2 – c2))z = 0. (4.1) The equations of EF and FD may be written down fro Equation (4.1) by cyclic change of x, y, z and a, b, c. It may now be verified that the equation of the line through A parallel to FD is (c2(a6 – a4(b2 + 3c2) + a2(b2 + 3c2)(c2 – b2) + b6 – 3b4c2 + 3b2c4 – c6))y + (a8 – 3a6(b2 + c2) + a4(3b4 + 2b2c2 + 3c4) + a2(b4 – c4(c2 – b2)))z = 0.
(4.2)
The equation of the line through B parallel to DE may be obtained from Equation (4.2) by cyclic change of x, y, z and a, b, c. These two lines and the line through C parallel to EF all pass through the point W with coordinates W(1/{b2(c2 + a2 – b2)}, 1/{c2(a2 + b2 – c2)}, 1/{a2(b2 + c2 – a2)}). (4.3) Similarly the line through A parallel to D'E', the line through B parallel to E'F' and the line through C parallel to F'D' all pass through the point W' with co-ordinates W'(1/{c2(a2 + b2 – c2)}, 1/{a2(b2 + c2 – a2)}), 1/{b2(c2 + a2 – b2)}). (4.4) The equation of BW is x/{a2(b2 + c2 – a2)} = z/{b2(c2 + a2 – b2)},
(4.5)
and the equation of CW' is x/ a2(b2 + c2 – a2)} = y/{c2(a2 + b2 – c2)}
(4.6)
These lines meet at L with co-ordinates L(a2(b2 + c2 – a2), c2(a2 + b2 – c2), b2(c2 + a2 – b2)).
(4.7)
Similarly M has co-ordinates M(c2(a2 + b2 – c2), b2(c2 + a2 – b2), a2(b2 + c2 – a2)),
(4.8)
and N has co-ordinates 4
N(b2(c2 + a2 – b2), a2(b2 + c2 – a2)), c2(a2 + b2 – c2)).
(4.9)
5. The conic through W, W', L, M, N The conic through W, W', L, M, N has the form of Equation (3.1) where u = 2b2c2(c2 + a2 – b2)(a2 + b2 – c2), v = 2c2a2(a2 + b2 – c2)(b2 + c2 – a2), w = 2a2b2(b2 + c2 – a2)(c2 + a2 – b2), f = – a4(b2 + c2 – a2)2, g = – b4(c2 + a2 – b2)2, h = – c4(a2 + b2 – c2)2.
(5.1) (5.2) (5.3) (5.4) (5.5) (5.6)
It may now be checked that this conic passes through O and K and also has centre X, the midpoint of OK. This conic we therefore term the twin conic of the Brocard circle. However, it is not merely the construction that mirrors that of the Brocard circle, but the fine detail also. Let us therefore look at the actual equation of the Brocard circle. Its form is again that of Equation (3.1) with u = 2b2c2, v = 2c2a2, w = 2a2b2, f = – a4, g = – b4, h = – c4. (5.7) As one can see one moves from the Brocard circle to its twin by replacing (a2, b2, c2) by (a2(b2 + c2 – a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)). In fact further than this, whereas in the Brocard circle the Brocard points Ω and Ω' have co-ordinates (1/b2, 1/c2, 1/a2) and (1/c2, 1/a2, 1/b2) respectively then W and W' have co-ordinates given by Equations (4.3) and (4.4) governed by the same replacement. The same applies to L, M, N. For example in the Brocard circle L has coordinates (a2, c2, b2) but in the twin L has co-ordinates (a2(b2 + c2 – a2), c2(a2 + b2 – c2), b2(c2 + a2 – b2)). 6. The conjugation For the above process to be termed a conjugation we have to be sure that if A goes to B in the conjugation then B goes to A. Suppose then that F = f(g + h – f), G = g(h + f – g), H = h(f + g – h), we must see what happens if we work out F(G + H – F), G(H + F – G) and H(F + G – H). In fact F(G + H – F) = f(g + h – f)(h + f – g)(f + g – h), G(H + F – G) = g(g + h – f)(h + f – g)(f + g – h), H(F + G – H) = h(g + h – f)(h + f – g)(f + g – h),
5
(6.1) (6.2) (6.3)
and since we are dealing with co-ordinates, the common factors may be eliminated leaving (f, g, h). And so in the analysis in Section 5 we see that O and K are conjugates. The work given in this article may be extended to other conics and their twin partners when one uses the construction started by drawing parallels. For example there will be an I-conic using the construction starting with the incentre I which will pass through the Mittenpunkt Mi, and there will be a twin Mi-conic starting with the Mittenpunkt which will pass through the incentre and both will have centres the midpoint of the segment IMi. They will both be seven point conics in which the seven points are related to each other by the conjugation (a, b, c) to (a(b + c – a), b(c + a – b), c(a + b – c)). It is very likely that this conjugation will have applications in contexts other than the one highlighted here.
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Article: CJB/2011/191
Three Circles and their Centres Christopher Bradley Abstract: Given a cyclic quadrilateral and a point P not on a side, lines may be drawn through P parallel to its sides each line meeting the adjacent sides in a point. The eight points created in this way determine two circles and the three circle centres and P form a parallelogram.
A A' D'
C'' O D Q
R
A'' B''
P
C
B' C' B
D''
Fig. 1 The three circles and their centres O, Q, R 1. The point P and the lines parallel to AB, BC, CD, DA through P We take the circle ABCD to be the unit circle centre the origin O. Co-ordinates of A are (x, y), where x = (1 – a2/(1+ a2), y = 2a/(1 + a2). (1.1) Points B, C, D have similar co-ordinates but using parameters b, c, d. 1
The equation of AB is then well known to be (1 – ab)x + (a + b)y = (1 + ab).
(1.2)
Equations of BC, CD, DA are similar but with appropriate change of parameters. We take P to have co-ordinates (k, 0) The equation of the line through P parallel to AB is clearly (1 – ab)x + ( a + b)y = (1 – ab)k.
(1.3)
The equations of the lines through P parallel to the other sides are similar but with appropriate change of parameters. 2. The four points A', B', C', D' and the circle A'B'C'D' The line through P parallel to AB meets AD at point A' whose co-ordinates are (x, y), where x = (a2(bk + d) + a(bd(k + 1) – k + 1) + b – dk)/{(1 + a2)(b – d), (2.1) 2 y = ((ab – 1)(ad(k + 1) – k + 1))/{(1 + a )(b – d). (2.2) The line through P parallel to AB meets BC at B' whose co-ordinates are (x, y), where x = (a(b2k + bc(k + 1) + 1) + b2c + b(1 – k) – ck)/{(1 + b2)(a – c)}, y = ((ab – 1)(bc(k + 1) – k + 1))/{(1 + b2)(a – c)}.
(2.3) (2.4)
The line through P parallel to CD meets BC at C' whose co-ordinates are(x, y), where x = (b(c2 + cd(k + 1) – k) + c2dk + c(1 – k) + d)/{(1 + c2)(d – b)}, y = ((1 – cd)(bc(k + 1) – k + 1))/{(1 + c2)(b – d)}.
(2.5) (2.6)
The line through P parallel to CD meets AD at D' whose co-ordinates are (x, y), where x = (a(cd(k + 1) + d2 – k) + c(d2k + 1) + d(1 – k))/{(1 + d2)(c – a)}, y = ((1 – cd)(ad(k + 1) – k + 1))/{(1 + d2)(a – c)}.
(2.7) (2.8)
The equation of the circle A'B'C'D' may now be obtained and is of the form x2 + y2 + 2gx + 2fy + h = 0,
(2.9)
where we find f = (a(b(c + d)(k + 1) + cd(k + 1) – k + 1) + b(cd(k + 1) – k + 1) + c(1 – k) + d(1 – k)) all divided by{2(a – c)(d – b)}, g = (ab(cd(k + 1) – k) – cdk + k – 1)/{(a – c)(b – d)}, h = (a(b(cd(k + 1)(k – 1) – k2) – d) – bc – cdk2 + k2 – 1)/{(a – c)(b – d)}.
(2.10) (2.11) (2.12)
2
3. The four points A'', B'', C'', D'' and the circle A''B''C''D'' The line through P parallel to BC meets AB at B'' whose co-ordinates are (x, y), where x = (a(b2 + bc(k + 1) – k) + b2ck + b(1 – k) + c)/{(1 + b2)(c – a)}, y = ((1 – bc)(ab(k + 1) – k + 1))/{(1 + b2)(a – c)}.
(3.1) (3.2)
The line through P parallel to BC meets CD at C'' whose co-ordinates are (x, y), where x = (b(c2k + cd(k + 1) + 1) + c2d + c(1 – k) – dk)/{(1 + c2)(b – d)}, y = ((bc – 1)(cd( k + 1) – k + 1))/{(1 + c2)(b – d)}
(3.3) (3.4)
The line through P parallel toad meets AB at A'' with co-ordinates (x, y), where x = (a2(b + dk) + a(bd(k + 1) – k + 1) – bk + d)/{(1 + a2)(d – b)}, y = ((ab(k + 1) – k + 1)(ad – 1))/{(1 + a2)(d – b)}.
(3.5) (3.6)
The line through P parallel to AD meets CD at D'' with co-ordinates (x, y), where x = (a(cd(k + 1) + d2k + 1) + c(d2 – k) + d(1 – k))/{(1 + d2)(a – c)}, y = ((ad – 1)(cd(k + 1) – k + 1))/{(1 + d2)(a – c)}.
(3.7) (3.8)
The equation of the circle A''B''C''D'' may now be obtained and it has the form of equation (2.9), where f = (a(b(c + d)(k + 1) + cd(k + 1) – k + 1) + b(cd(k + 1) – k + 1) + c(1 – k) + d(1 – k)) all divided by {2(a – c)(b – d)}, (3.9) g = (ad(bc(k + 1) – k) – bck + k – 1)/{(a – c)(b – d)}, (3.10) h = (a(b(cd(k + 1)(k – 1) – 1) – dk2) – bck2 – cd + k2 – 1)/{(a – c)(b – d)}. (3.11) 4. The circle centres The centres of the circles in Sections 2 and 3 have centres (– g, – f ). It may now be checked that RO and PQ are equal and parallel, their joint displacement vector being (x, y), where x = (ad(bc(k + 1) – k) – bck + k – 1)/{(a – c)(d – b)}, (4.1) y = (a(b(c + d)(k + 1) + cd(k + 1) – k + 1) + b(cd(k + 1) – k + 1) + (1 – k)(c + d)) all divided by {2(a – c)(b – d)}. (4.2)
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Article: CJB/2011/192
Generalization of the Triplicate Ratio Circle Christopher Bradley Abstract: Given a triangle ABC a point P not on its sides is selected and lines parallel to AB and AC are drawn through P to meet the sides in four distinct points. The position of P is determined when these four points are concyclic.
A
V M
P O
B
X
K
U
L
C
Fig. 1 A Special Circle centre on the Brocard axis 1. The points L, M, U, V The point P is chosen not on the sides of ABC with general co-ordinates (f, g, h), f, g, h ≠ 0. The line through P parallel to AB has equation hx + hy – (f + g)z = 0 (1.1) 1
It meets BC at the point L with co-ordinates L(0, f + g, h). It also meets CA at the point M with co-ordinates M(f + g, 0, h). The line through P parallel to AC has equation gx – (h + f)y + gz = 0. (1.2) It meets BC at the point U with co-ordinates U(0, g, h + f). It also meets AB at the point V with co-ordinates (h + f, g, 0). 2. The condition for L, M, U, V to be concyclic The equation of a circle in areal co-ordinates is of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
A circle passes through L, M, U when u = (h(a2g – b2(f + g + h))/(f + g + h)2, v = – (a2h(h + f))/(f + g + h)2, w= – (a2g(f + g))/(f + g + h)2.
(2.2) (2.3) (2.4)
It also passes through V, assuming U, V, L, M are distinct, when c2g = b2h.
(2.5)
In geometrical terms this means that P must lie on the line AK, where K is the symmedian point. It may be checked that the centre of the circle lies on the Brocard axis.
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Article: CJB/2011/193
A Two Circle Problem Involving the GK axis Christopher Bradley Abstract: Through a point X lines are drawn parallel to the sides of ABC. From the resulting six points two circles are drawn and the condition on X for their common chord to pass through X is determined, the result being that X must lie on the GK axis.
A
N
Q
M
U R
B
G X
L
K V
P
C
Fig. 1 1. The six points L, M, N, P, Q, R We take X to have co-ordinates (f, g, h), where f, g, h ≠ 0. The line through X parallel to AB has equation h(x + y) – (f + g)z = 0.
1
(1.1)
It meets BC at L with co-ordinates L(0, f + g, h). It meets CA at Q with co-ordinates Q(f + g. 0. h). The line through X parallel to BC has equation (g + h)x – f(y + z) = 0.
(1.2)
It meets AB at R with co-ordinates R(f, g + h, 0). It meets CA at M with co-ordinates M(f, 0, g + h). The line through X parallel to CA has equation gx – (h + f)y + gz = 0.
(1.3)
It meets BC at P with co-ordinates (0, g, h + f). It meets AB at N with co-ordinates N(h + f, g, 0). 2. The circles LMN and PQR The circles have equations of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(2.1)
For circle LMN, putting k = (f + g + h)2(gh + hf + fg) we find u= (1/k)(g(g + h)(a2h(f + g) – b2hf – c2(f + g)(h + f)), v = (1/k)(h(h + f)(f(b2(g + h) – c2g) – a2(f + g)(g + h)), w = – (1/k)(f(f + g)(a2gh + (h + f)(b2(g + h) – c2g)).
(2.2) (2.3) (2.4)
For circle PQR, with k as before, we find u = (1/k)(h(g + h)(a2g(h + f) – b2(f + g)(h + f) – c2fg)), v = (1/k)(f(h + f)((f + g)(b2h – c2(g + h)) – a2gh)), w = – (1/k)(g(f + g)(a2(h + f)(g + h) + f(b2h – c2(g + h)))).
(2.5) (2.6) (2.7)
3. The common chord The common chord of two circles S1 and S2 has the form (u1 – u2)x + (v1 – v2)y + (w1 – w2)z = 0.
(3.1)
Using values of u, v. w from Section 2 we find the equation of the common chord of LMN and PQR is x(g + h)(a2gh(g – h) + b2h(f2 + hf + gh) – c2g(f2 + fg + gh)) – y(h + f)(a2h(hf + g2 + gh) + f(b2h(f – h) – c2(f(g + h) + g2))) + z(f + g)(a2g(fg + gh + h2) – f(b2(f(g + h) + h2) + c2g(g – f))) = 0. (3.2) 2
The common chord passes through X(f, g, h) if, and only if, a2(g – h) + b2(h – f) + c2(f – g) = 0 That is if, and only if, X lies on the line through G(1, 1, 1) and K(a2, b2, c2),
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(3.3)
Article: CJB/2011/194
The KG Nine-Point Conic Christopher Bradley Abstract: The properties of a Nine-Point Conic are exhibited using Cevian point K (as well as the centroid G). The symmedian point Km of the medial triangle (X141) is shown to lie on the KG axis, the midpoint of KKm being the centre X of the nine-point conic.
A
W N
M X K
Km
B
V
G
L
U
C
Fig. 1 1. The KG Conic This is the conic that passes through the feet of the Cevians of G, the centroid, and K the symmedian point. 1
Its equation is of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(1.1)
u = 2b2c2, v = 2c2a2, w = 2a2b2,
(1.2)
f = – a2(b2 + c2), g = – b2(c2 + a2), h = – c2(a2 + b2).
(1.3)
where we find
and
Its centre is easily found to be the point X with co-ordinates (2a2 + b2 + c2, 2b2 + c2 + a2, 2c2 + a2 + b2).
(1.4)
2. The position of the centre The axis GK has equation (b2 – c2)x + (c2 – a2)y + (a2 – b2)z = 0.
(2.1)
X clearly lies on this line, as does X141 = Km, the symmedian point of the medial triangle whose co-ordinates are (b2 + c2, c2 + a2, a2 + b2). Indeed X is the midpoint of KKm. Since K(a2, b2, c2) can be changed to any triple P(f, g, h) it follows that the nine-point conic based on P and G always has its centre midway between P and Pm(g + h, h + f, f + g), the P-point of the medial triangle.
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2
Article: CJB/2011/195
The Orthocentre Circle of a Cyclic Quadrilateral Christopher Bradley Abstract: With ABCD a cyclic quadrilateral perpendiculars are drawn from A to BC and to CD meeting those sides at L and N respectively and perpendiculars are drawn from B to DA and to CD meeting those sides at Z and X respectively. Further the perpendiculars from C to AB and DA are drawn meeting those sides at M and P respectively and finally perpendiculars are drawn from D to AB and BC meeting those sides at Y and W respectively. Point A' = BZ^DY, point B' = CM^AL, point C' = BX^DW and point D' = AN^CP. It is found that A', B', C', D' are concyclic and The circle we term the orthocentre circle as it mirrors the construction of the orthocentre in a triangle. Nine other circles arise out of the figure and their properties are investigated.
V'
A
E' Z
A''
N D'
Q' D''
R' D
O Y
A' U
F
P S' W
R
C
B'
S
T'
O' M B''
T
L C''
Q
B E
X
V
1
C'
1. The twelve key points Cartesian co-ordinates are used throughout with circle ABCD the unit circle centre O the origin. Point A has co-ordinates x = (1 – a2)/(1 + a2), y = 2a/(1 + a2), with B, C, D similar but with parameters b, c, d. The perpendicular from B to AD has equation (1 + b2){(a + d)x – (1 – ad)y} = – {a(b2 – 2bd – 1) + b2d + 2b – d}.
(1.1)
This line meets AD at the point Z with co-ordinates (x, y), where k = (1 + a2)(1 + b2)(1 + d2) and x = – (1/k)(a2(b2(1 + d2) – 2bd + d2 – 1) + 2a(b2d + b(1 – d2) – d) + b2(d2 – 1) + 2bd – d2 – 1), y = (1/k)(2(a2d(bd + 1) + a((b – d)2 + b(bd + 1))). (1.2) Note that these expressions are invariant under the exchange of a and d. The co-ordinates of Y may be obtained from those of Z by exchange of b and d. The co-ordinates of M may be obtained from those of Y by replacing d with c. The co-ordinates of P may be found from those of Z by replacing b with c. The co-ordinates of X may be found from those of Z by replacing a with c. The co-ordinates of N may be obtained from those of X by replacing b with a. The co-ordinates of W may be obtained from those of Y by replacing a with c. The co-ordinates of L may be found from those of W by replacing d with a. The perpendicular from B to AD has equation (1.1) and the perpendicular from D to AB has an equation obtained from (1.1) by exchange of b and d. These two lines meet at the point A' with co-ordinates (x, y), where x = – (1/k)(a2(b2(3d2 + 1) + d2 – 1) + b2(d2 – 1) – d2 – 3), y = (2/k)(a2(b2d + b(d2 + 1) + d) + a(b2 + 1)(d2 + 1) + b2d + b(d2 + 1) + d). (1.3) The co-ordinates of B', C', D' may be obtained from those of A' by cyclic change of a, b, c, d. 2. The circles Circles have an equation of the form x2 + y2 + 2gx + 2fy + h = 0, where the centre has co-ordinates (– g, – f) and the radius r = √(g2 + f2 – h).
2
(2.1)
Substituting in (2.1) the co-ordinates of A', B', C', D' we find the circle A'B'C'D' has values of f, g, h as follows, where j = {(1 + a2)(1 + b2)(1 + c2)(1 + d2)}, f = – (2/j)(a2(b2(c2d + c(d2 + 1) + d) + b(c2 + 1)(d2 + 1) + c2d + c(d2 + 1) + d) + a(b2 + 1)(c2 + 1)(d2 + 1) + b2(c2d + c(d2 + 1) + d) + b(c2 + 1)(d2 + 1) + c2d + c(d2 + 1) + d), g = (2/j)(a2(b2(c2(2d2 + 1) + d2) + c2d2 – 1) + b2(c2d2 – 1) – c2 – d2 – 2), (2.2) 2 2 2 2 2 2 2 2 2 h = (1/j)(a (b (3c (5d + 1) + 8cd + 3d – 1) + 8b(c d + c(d + 1) + d) + c (3d – 1) + 8cd – d2 + 3) + 8a(b2(c2d + c(d2 + 1) + d) + b(c2(d2+ 1) + d2 + 1) + c2d + c(d2 + 1) + d) + b2(c2(3d2 – 1) + 8cd – d2 + 3) + 8b(c2d + c(d2 + 1) + d) + c2(3 – d2) + 8cd + 3d2 + 15). It may be checked that the radius of this circle is 1, the same as that of circle ABCD. It may now be checked that the co-ordinates of O' are (x, y), where x = (1 – a2)/(1 + a2) + (1 – b2)/(1 + b2) + (1 – c2/(1 + c2) + (1 – d2)(1 + d2), y = 2a/(1 + a2) + 2b/(1 + b2) + 2c/(1 + c2) + 2d/(1 + d2). (2.3) This is a highly significant result, for in vector terms we have OO' = OA + OB + OC + OD,
(2.4)
which justifies (by comparison with a triangle) us calling O' the orthocentre and circle A'B'C'D' the orthocentre circle. It may now be shown that L, M, B, B' are concyclic with radius = √((1 + ac)2/{(1 + a2)(1 + c2)}) and N, P, D, D' are concyclic with the same radius Again Y, Z, A, A' are concyclic with radius = √((1 + bd)2/{(1 + b2)(1 + d2)}) and X, W, C, C' are concyclic with the same radius. The centres of these four circles are labelled B'', D'', A'', C'' respectively. It may now be shown that points A, C, B', D' are concyclic with radius 1 and similarly B, D, A', C' are concyclic also with radius 1. Further it may be shown that A, C, L, M, N, P are concyclic with radius = √((1– ac)2/{(1 + a2)(1 + c2)}) and that B, D, X, Y, Z, W are concyclic with radius = √((1– bd)2/{(1 + b2)(1 + d2)}). The co-ordinates of B'', D'', A'', C'' may be obtained from the equations of the four circles in question and it may now be shown that the figure A''B''C''D'' is a parallelogram. Q', the centre of circle ACB'D' has co-ordinates (2(1 – a2c2)/{(1 + a2)(1 + c2)}, 2(1 + ac)ac/{1 + a2)(1 + c2)}), And Q has similar co-ordinates with b, d replacing a, c. 3
(2.5)
The midpoint of QQ' is U with co-ordinates half those of O', showing U is the midpoint of OO'. Carrying the analogy with a triangle further this shows U is the analogue of the nine-point centre. There are actually two circles centre U, namely LPWZ and MNXY but neither have radius ½ so it would be stretching things too far to say they are analogues of the nine-point circle. The point U is the intersection of the diagonals of parallelogram A''B''C''D'' and as can be seen from the figure lies on many other lines, such as the common chord of circles ABCD and A'B'C'D'. It is left to the reader to check that XYZW and LMNP are both circles, centres R and R' respectively which are the midpoints of the diagonals BD and AC respectively.
Flat 4 Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
4
Article: CJB/2011/196
The Two Central Lines in a Cyclic Quadrilateral Christopher Bradley Abstract: There are five main central points in a cyclic quadrilateral: the circumcentre O, the centroid F, the centre of mass G, the intersection of the diagonals E and the intersection of the lines joining the midpoints of opposite sides T. It is proved in this article that GTE is a straight line and that OTF is a straight line. Furthermore ET = 3TG and OT = 3TF and so GF is parallel to OE. (The anticentre also lies on OTF.) D A
L
S
E K
P
F T
R G
O N
M
B
Q
C
Fig.1 The central lines of a cyclic quadrilateral 1
1. The centroid F We use areal co-ordinates with ABC the triangle of reference and D the general point on the circumcircle with co-ordinates (x, y, z), where x = – a2t(1 – t), y = b2(1 – t), z = c2t. (1.1) We label the midpoints of AB, BC, CD, DA to be P, Q, R, S respectively. The lines PR QS meet at a point T with normalized co-ordinates (x, y, z), where x = (2a2t(t – 1) + b2(1 – t) + c2t)/{4(a2t(t – 1) + b2(1 – t) + c2t)}, (1.2) 2 2 2 2 2 2 y = (a t(t – 1) + 2b (1 – t) + c t)/{4(a t(t – 1) + b (1 – t) + c t)}, (1.3) 2 2 2 2 2 2 z = (a t(t – 1) + b (1 – t) + 2c t)/{4(a t(t – 1) + b (1 – t) + c t)}. (1.4) We label the centroids of triangles DAB. ABC, BCD, CDA to be L, M, N, P respectively. Their normalized co-ordinates are: L: x = (2a2t(t – 1) + b2(1 – t) + c2t)/{3(a2t(t – 1) + b2(1 – t) + c2t)}, y = (a2t(t – 1) + 2b2(1 – t) + c2t)/{3(a2t(t – 1) + b2(1 – t) + c2t)}, (1.5) 2 2 2 2 z = c t/{3(a t(t – 1) + b (1 – t) + c t)}. M: x = y = z = 1/3. (1.6) 2 2 2 2 N: x = a t(t – 1)/{3(a t(t – 1) + b (1 – t) + c t)}, y = (a2t(t – 1) + 2b2(1 – t) + c2t)/{3(a2t(t – 1) + b2(1 – t) + c2t)}, (1.7) 2 2 2 2 2 2 z = (a t(t – 1) + b (1 – t) + 2c t)/{3(a t(t – 1) + b (1 – t) + c t)}. P: x = (2a2t(t – 1) + b2(1 – t) + c2t)/{3(a2t(t – 1) + b2(1 – t) + c2t)}, y = b2(1 – t)/{3(a2t(t – 1) + b2(1 – t) + c2t)}, (1.8) 2 2 2 2 2 2 z = (a t(t – 1) + b (1 – t) + 2c t)/{3(a t(t – 1) + b (1 – t) + c t)}. The equation of the circle LMNP is of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0,
(1.9)
where we find k = 9(a2t(t – 1) + b2(1 – t) + c2t), and u = (1/k)(a4t(t – 1) – a2(b2(t – 1)(t + 2) + c2t(t – 3)) + 2(b4(t – 1) – 2b2c2 – c4t), v = – (1/k)(2a4t(t – 1) + a2(4c2t2 – b2(t – 1)(2t + 1)) + b4(t – 1) + b2c2(1 – 3t) + 2c4t), w = – (1/k)(2a4t(t – 1) + a2(4b2(t – 1)2 + c2t(3 – 2t)) + (c2 – b2)(2b2(t – 1) – c2t)).
(1.10) (1.11) (1.12)
The centre of this circle is the centroid F which has co-ordinates x = – (a4 – a2(b2 + c2 + 2u – v – w) + (b2 – c2)(v – w)), y = a2(b2 + w – u) – b4 + b2(c2 – u + 2v – w) + c2(u – w), z = a2(c2 – u + v) + b2(c2 + u – v) – c4 – c2(u + v – 2w). Here u, v, w are given by Equations (1.10) – (1.12) 2
(1.13)
With circumcentre O it is now straightforward to show that O, T and F are collinear and OT = 3TF. A final note is worth making. If you take the midpoints of the sides P, Q, R, S and drop perpendiculars on to opposite sides they concur at a point that is known by the horrible name ‘the anticentre Ă’ and it satisfies OT = TĂ. If a, b, c, d are the vector positions of A, B, C, D with respect to O then OT = (1/4)(a + b + c + d) and OF = (1/3)(a + b + c + d) and OĂ = (1/2)(a + b + c + d). It is true that the whole of this Section would have been more easily established using vectors, but this would not be the case for the second central line GTE. 2. The centre of mass G and the line GTE L, M, N and P are the centres of mass of triangles DAB, ABC, BCD and CDA respectively. It follows that the centre of mass G of ABCD lies on the line LN and also on the line MP and hence G is the point of intersection of LN and MP. The equation of LN is (a2t(t – 1) + 2b2(1 – t) + c2t)x – (2a2t(t – 1) + b2(1 – t) + 2c2t)y + (a2t(t – 1) + 2b2(1 – t) + c2t)z = 0 (2.1) The equation of MP is (a2(t – 1) + 2c2)x +(a2(t – 1) – c2)y – (2a2(t – 1) + c2)z = 0. Lines LN and MP meet at the centre of mass G with co-ordinates x= a4t(t – 1)2 + 2a2c2t(t – 1) – c2b2(t – 1) + c4t, y = (a2t(t – 1) + 2b2(1 – t) + c2t)(a2(t – 1) + c2), z = a4t(t – 1)2 + a2(1 – t)(b2(t – 1) – 2c2t) + c4t. Line AC has equation y = 0 and line BD has equation c2x + a2(1 – t)z = 0.
(2.2)
(2.3)
(2.4)
These two lines meet at E with co-ordinates E(a2(t – 1), 0, c2). It may now be shown that GTE is a straight line and that ET = 3TG. Unit masses at A, B, C, D do, of course have centre of mass at T. The distinction between these various points is sometimes quite unclear in the literature. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article: CJB/2011/197
When Two Pairs of Diagonals are Concurrent Christopher Bradley Abstract: If ABCD is a cyclic quadrilateral and the tangents at A, B, C, D form a quadrilateral PQRS, the diagonals AC, BD and PR, QS all pass through a given point. Here we identify this point.
P
A
S B
O
T
D
R C
Q Fig. 1 Two quadrilaterals with the same diagonal point 1. The cyclic quadrilateral ABCD and the tangents at A, B, C, D We use areal co-ordinates throughout with ABC the triangle of reference and circle ABCD having equation 1
a2yz + b2zx + c2xy = 0 We take the co-ordinates of D to be (x, y, z), where x = – a2t(1 – t), y = b2(1 – t), z = c2t. The equations of the tangents at A, B, C, D are A: c2y + b2z = 0, B: c2x + a2z = 0, C: b2x + a2y = 0, D: x/a2 + t2y/b2 + (1 – t)2z/c2 = 0.
(1.1)
(1.2)
(1.3)
2. The points P, Q, R, S, T The tangents at A and B meet at P with co-ordinates P(a2, b2, – c2). The tangents at B and C meet at Q with co-ordinates Q(– a2, b2, c2). The tangents at C and D meet at R with co-ordinates R(– a2(1 – t), b2(1 – t), c2(1 + t)). The tangents at D anA meet at S with co-ordinates S(a2(2t – 1), – b2, c2). AC has equation y = 0, BD has equation c2tx + a2t(1 – t)z = 0 so that T = AC^BD has coordinates x = a2(t – 1), y = 0, z = c2. (2.1) The equation of PR is x/a2 – ty/b2 + (1 – t)z/c2 = 0
(2.2)
x/a2 + ty/b2 + (1 – t)z/c2 = 0.
(2.3)
The equation of QS is
It is now easily checked that PR^QS = T also. Opposite sides of both quadrilaterals intersect on the polar of T with respect to ABCD.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/2011/198
The Two Cevians’ Perspective Christopher Bradley
A
W
T
F
R P L M
B
D
S
E V
O Q N
U
C
K
Fig.1 Abstract: In a triangle Cevians are drawn through P and Q. The points L, M, N, R, S, T are defined as follows: L = BQ^CP, M = BQ^AP, N = AQ^CP, R = CQ^BP, S = AQ^BP, T = CQ^AP. It is proved that triangles LMN and RST are in perspective. 1. The Conic through the feet of the Cevians Let P and Q have co-ordinates P(l, m, n), Q(r, s, t). Then the feet of the Cevians are D(0, m, n), E(l, 0, n), F(l, m, 0), U(0, s, t), V(r, 0, t), W(r, s, 0). The equation of the conic through these six points is well known to be
ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(1.1)
where u = 2mnst, v = 2nltr, w = 2lmrs, f = – lr(mt + ns), g = – ms(lt + nr), h = – nt(ls + mr). 2. The Perspective The Cevians themselves have equations as follows: AP: mz = ny, AQ: sz = ty, BP: nx = lz, BQ: tx = rz, CP: ly = mx, CQ: ry = sx.
(2.1)
The co-ordinates of the six points are: L = BQ^CP: x = lr, y = mr, z = lt, M = BQ^AP: x = nr, y = mt, z = nt, N = AQ^CP: x = ls, y = ms, z = mt, R = CQ^BP: x = lr, y = ls, z = nr, S = AQ^BP: x = lt, y = ns, z = nt, T = CQ^AP: x = mr, y = ms, z = ns,
(2.2) (2.3) (2.4) (2.5) (2.6) (2.7)
The equation of LR is (l2st – mnr2)x – lr(lt – nr)y – lr(ls – mr)z = 0.
(2.8)
nt(ns – mt)x + nt(nr – lt)y + (lmt2 – n2rs)z = 0.
(2.9)
ms(mt – ns)x +(nls2 – m2tr)y + ms(mr – ls)z = 0.
(2.10)
The equation of MS is
The equation of NT is
It may now be checked that all three lines meet at a point O with co-ordinates x = lr(mt + ns), y = ms(lt + nr), z = nt(ls + mr).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
(2.11)
Article: CJB/2011/199
Three Circles with collinear Centres Christopher Bradley
D
S A
U
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J E
R
O
V P
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C K Q
B W
Fig. 1 Abstract: With ABCD a cyclic quadrilateral centre O and midpoints of sides PQRS, lines PO, QO, RO, SO meet the opposite sides at T, U, V, W respectively. Circles SUQW and TVPR have centres F and E respectively. It is proved that FOE is a straight line. Some other obvious results are mentioned.
1
1. Introduction With ABCD a cyclic quadrilateral centre O and midpoints of sides P, Q, R, S, lines PO, QO, RO, SO meet the opposite sides at T, U, V, W respectively. Circles SUQW and TVPR have centres F and E respectively. It is proved that FOE is a straight line. Circles ASOP, DROS, CQOR, BPOQ have centres J, K, L, M respectively. Circle JKLM has centre O and half the radius of circle ABCD. Work is carried out using Cartesian co-ordinates. 2. Points A, B, C, D, P, Q, R, S We take circle ABCD to have origin O and radius 1, with the co-ordinates of A to be (x. y), where x = (1 – a2)/(1 + a2), y = 2a/(1 + a2). (2.1) Points B, C, D have similar co-ordinates with parameters b, c, d respectively. Point P is the midpoint of AB and therefore has co-ordinates x = {(1 – ab)(1 + ab)}/{(1 + a2)(1 + b2)}, y = {(a + b)(1 + ab)}/{(1 + a2)(1 + b2)}.
(2.2)
Co-ordinates of Q, R, S are similar, with appropriate change of parameters b, c and c, d and d, a respectively. 3. Lines OP, OQ, OR, OS AB, BC, CD, DA Line OP has equation (a + b)x + (ab – 1)y = 0.
(3.1)
Lines OQ, OR, OS have similar equations but with parameters b, c and c, d and d, a. Line AB has equation (1 – ab)x + (a + b)y = 1 + ab.
(3.2)
Lines BC, CD, DA have similar equations bur with parameters b, c and c, d and d, a. 4. Points T, U, V, W Point T = OP^CD has co-ordinates x = (1 – ab)(1 + cd)/(abcd – ab + ac + ad + bc + bd – cd + 1), y = (a + b)(1 + cd)/(abcd – ab + ac + ad + bc + bd – cd + 1). 2
(4.1) (4.2)
Points U, V, W have similar co-ordinates with a, b, c, d replaced in turn by b, c, d, a and c, d, a, b and d, a, b, c respectively. 5. Circles USQW and VTPR Circles have equations of the form x2 + y2 + 2gx + 2fy + k = 0.
(5.1)
Circle USQW has f = – (abc + abd + acd + bcd + a + b + c + d)/2(abcd + ab + ac – ad + bd – bc + cd + 1), g = (abcd – 1)/(abcd + ab + ac – ad + bd – bc + cd + 1), k = (1 + ad)(1 + bc)/(abcd + ab + ac – ad + bd – bc + cd + 1).
(5.2) (5.3) (5.4)
Circle VTPR has f' = – (abc + abd + acd + bcd + a + b + c + d)/2(abcd – ab + ac + ad + bc + bd – cd + 1), (5.5) g' = (abcd – 1)/(abcd – ab + ac + ad + bc + bd – cd + 1), (5.6) k' = (1 + ab)(1 + cd)/(abcd – ab + ac + ad + bc + bd – cd + 1). (5.7) The centre of circles of the form (5.1) have co-ordinates (– g, – f). Circle USQW has centre F and circle VTPR has centre E and because fg' = f'g it follows that F, E, O are collinear. 6. Circles ASOP, BPOQ, CQOR, DROS It is easily shown that the centres J, K, L, M of these circles are such that OJ = JA, OK = KB, OL = LC, OM = MD and hence that the circle JKLM has centre O and radius half that of circle ABCD.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/ 2011/200
Problem on Quadrilateral with an Incircle Christopher Bradley Abstract: Given a quadrilateral with an incircle there are vertices A, B, C, D and points of contact P, Q, R, S. Lines AQ, AR, CP, CS are drawn intersecting in K and M. Lines BS, BR, DP, DQ are drawn intersecting in N and L. It is proved that K, M lie on BD and N, L lie on AC.
A
S' S
P' N'
P
N D M K
T O
B
M' K' L R
L'
Q
R'
Q'
C
Fig. 1
1
1. Introduction Areal co-ordinates are used throughout with PQR the triangle of reference and so the equation of the circle involved is a2yz + b2zx + c2xy = 0. (1.1) The point S may be taken to have co-ordinates (x, y, z), where x = a2t(1 – t), y = b2(1 – t), z = c2t.
(1.2)
The equation of the tangents at P, Q, R, S are as follows: P: c2y + b2z = 0, Q: c2x + a2z = 0, R: b2x + a2y = 0, S: x/a2 + t2y/b2 + (1 – t)2z/c2 = 0.
(1.3)
We now give the co-ordinates of A, B, C, D. A is the intersection of the tangents at P and S. So A has co-ordinates x = a2(2t – 1), y = – b2, z = c2. B is the intersection of the tangents at P and Q. So B has co-ordinates x = a2, y = b2, z = – c2. C is the intersection of tangents at Q and R. So C has co-ordinates x = – a2, y = b2, z = c2. D is the intersection of tangents at R and S. So D has co-ordinates x = a2(t – 1), y = b2(1 – t), z = (1 + t)c2. 2. The eight lines AQ, AR, CP, CS, BR, BS, DP, DQ The equations of these eight lines are AQ: (2t – 1)a2z – c2x = 0, AR: (1 – 2t)a2y – b2x = 0, CP: c2y – b2z = 0, CS: (2t – 1)x/a2 + t2y/b2 – (t – 1)2z/c2 = 0. BR: b2x – a2y = 0, BS: x/a2 – t2y/b2 – (t – 1)2z/c2 = 0, DP: b2(1 – t)z – c2(t + 1)y = 0, DQ: c2(1 + t)x + a2(1 – t)z = 0.
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6) (2.7) (2.8)
3. The points K, L, M, N and the lines on which they lie The point K = AQ^CP has co-ordinates x = (2t – 1)a2, y = b2, z = c2. The point M = AR^CS has co-ordinates x = a2(t – 1)(2t – 1), y = b2(1 – t), z = c2(3t – 1). The point N = BS^DP has co-ordinates x = a2(t – 1)(2t + 1), y = b2(1 – t), z = c2(t + 1). The point L = BR^DQ has co-ordinates x = a2(t – 1), y = b2(t – 1), z = c2(1 + t). The equation of the diagonal AC is x/a2 + ty/b2 + (1 – t)z/c2 = 0. The equation of the diagonal BD is 2
(3.1)
x/a2 – ty/b2 – (t – 1)z/c2 = 0.
(3.2)
It is now easy to check that K lies on BD, M lies on BD, N lies on AC and L lies on AC. Note that if the analysis is carried out with the midpoints of sides rather than points of contact, the resulting points K', L', M', N' are such that K'M' and KM are parallel and N'L' is parallel to NL. We do not prove this but Cabri verifies it.
Flat 4, Terrill Court, 12- 14 Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/201
All about a Cyclic Quadrilateral and its cousin Christopher Bradley
W
E V D'
R
D
A U
A' S
C'
Q P O
F B B'
C
Fig. 1 A Cyclic Quadrilateral ABCD and Associated Circles Abstract: Given a Cyclic Quadrilateral ABCD with circumcentre O and diagonals AC, BD, EF then the midpoints of the diagonals P, Q, R respectively are well known to be collinear. The points U, V, W are the intersections of the diagonals. It is first shown that circle UPQ passes through O and that if the line RU meets this circle at S, then the following circles all pass through S. These circles are VRP, WRQ, UVW, EFO, ACR, BDR. If the circles WVA, WVB, 1
WVC, WVD meet the circle ABCD at points A', B', C', D' then RAA', RBB', RCC', RDD' are all straight lines. It is not unreasonable to say that S is the most important point in the figure. 1. Introduction We use areal co-ordinates throughout with ABC the triangle of reference and D the point with co-ordinates x = – a2t(1 – t), y = b2(1 – t), z = c2t, which is a general point of circle ABC with equation a2yz + b2zx + c2xy = 0. (1.1) We first obtain the key equations of lines and diagonal points. The lines BC, CA, AB have equations x = 0, y = 0, z = 0 respectively. CD has equation b2x + a2ty = 0.
(1.2)
The line DA has equation b2(1 – t)z = c2ty.
(1.3)
And the diagonal BD has equation c2x + a2(1 – t)z = 0.
(1.4)
Now for the co-ordinates of some key points: U = AC^BD: x = a2(t – 1), y = 0, z = c2. E = AB^CD: x = a2t, y = – b2, z = 0. F = AD^BC: x = 0, y = b2(1 – t), z = c2t.
(1.5) (1.6) (1.7)
The remaining diagonal EF has equation x/a2 + ty/b2 + (t – 1)z/c2 = 0.
(1.8)
2. The midpoints of the diagonals P, Q, R and the line on which they lie The midpoint P of the diagonal AC obviously has co-ordinates x = 1, y = 0, z = 1. The midpoint Q of diagonal BD, after some calculation, turns out to have co-ordinates Q: x = (1/k)a2t(t – 1), y = (1/k)(a2t(t – 1) + 2b2(1 – t) + c2t), z = (1/k)c2t, where k = 2(a2t(t – 1) + b2(1 – t) + c2t).
(2.1) (2.2)
The midpoint R of the diagonal EF has co-ordinates R: x = a2t(b2(t – 1) – c2t), y = b2(a2t(t – 1) + 2b2(1 – t) + c2t), z = c2t(b2 – a2t).
(2.3)
It may now be checked that P, Q, R lie on the midpoint line with equation 2
(a2t(t – 1) + 2b2(1 – t) + c2t)(x – z) + t(c2 – a2(t – 1))y = 0. The diagonal points V = EF^AC and W = EF^BD have co-ordinates V: x = a2(1 – t), y = 0, z = c2. W: x = a2t(t – 1), y = 2b2(1 – t), z = c2t.
(2.4)
(2.5) (2.6)
3. The circles through the point S Using areal co-ordinates the equations of all circles are of the form a2yz + b2zx + c2xy + (ux + vy + wz)(x + y + z) = 0.
(3.1)
Hence in dealing with a circle it is sufficient to specify its values of u, v, w. Of course to check that a point such as S lies on a circle one has to have the full equation of the circle. This full equation, of course, has to be stored in the algebra software being used. Circle UPQ has u = – (1/j)b2c2, v = – (1/j)c2a2t, w = (1/j)a2b2(1 – t), where j = 2(c2 + a2(t – 1).
(3.2) (3.3)
From its equation it is now easy to check that the circumcentre O lies on circle UPQ. We choose to define S as the intersection of circle UPQ with the line RU. First we find the equation of RU, which is (a2t(t – 1) + 2b2(1 – t) + c2t)b2c2x – a2c2t2(a2(t – 1) – c2)y – a2b2(t – 1)(a2t(t – 1) + 2b2(1 – t) + c2t) = 0. (3.4) Unfortunately the point S = line RU^ circle UPQO has extremely cumbersome expressions for its co-ordinates which are as follows: x = a2t(t – 1)(a6t2(t – 1)2 + a4t(1 – t)(2b2(t – 1) – c2t) – a2t(b4(t – 1)2 + 2b2c2t(1 – t) + c4t(t + 1)) + 2b6(t – 1)2 + 3b4c2t(1 – t) + 2b2c4t + c6t2), (3.5) 2 2 2 2 2 2 2 2 y = 2b (1 – t)(a t – b )(a t(t – 1) + 2b (1 – t) + c t)(b (t – 1) – c t), (3.6) 2 6 2 2 4 2 2 2 2 4 2 2 z = c t(a t (t – 1) + a t(t – 1)(2b (t – 1) + c t(1 – 2t)) + a t(1 – t)(3b (t – 1) – 2b c t – c4t) + (2b2(t – 1) – c2t)(b4(t – 1) – c4t)). (3.7) Circle UVW has u = b2c4/(a4(t – 1)2 – c4), v = a2c2t(a2t(t – 1) + b2(1 – t) + c2t)/((a2t(t – 1) + 2b2(1 – t) + c2t)(a2(t – 1) + c2)), w = a4b2(t – 1)2/(c4 – a4(t – 1)2). It may now be shown that S lies on the circle UVW. 3
(3.8) (3.9) (3.10)
Circle RVP has u = b2c2/(2(a2(t – 1) – c2)), v = c2a2t2(c2 + a2 – b2)/(2(a2t – b2)(b2(t – 1) – c2t)), w = a2b2(1 – t)/(2(a2(t – 1) – c2)).
(3.11) (3.12) (3.13)
It may now be shown that S lies on the circle RVP. Circle RWQ has u = b2c2(a2t + b2(t – 2) – c2t)/(2(a2t – b2)(b2(t – 1) – c2t)), v = – c2a2t2/(2(a2t(t – 1) + 2b2(1 – t) + c2t)), w = a2b2(1 – t)(a2t + b2(t – 1) – c2t)/(2(a2t – b2)(b2(t – 1) – c2t)).
(3.14) (3.15) (3.16)
It may now be shown that S lies on circle RWQ. Circle BDR has u = (b2c2(a4t(t – 1) + 2a2b2(1 – t) + b4(t – 1) + 2b2c2(1 – t) + c4t))/m where m = 2(a2t – b2)(a2(t – 1) – c2)(b2(t – 1) – c2t), v = 0, w = (a2b2(1 – t)(a4t(t – 1) + 2a2b2(1 – t) + b4(t – 1) + 2b2c2(1 – t) + c4t))/m.
(3.17) (3.18) (3.19)
It may now be shown that S lies on circle BDR. Circle ACR has u = 0, v = a2c2t2(a4t(t – 1) + 2a2b2(1 – t) + b4(t – 1) + 2b2c2(1 – t) + c4t) all divided by 2(a4t2(t – 1) + a2t(c2t – 3b2(t – 1)) + b2(2b2(t – 1) – c2t))(b2(t – 1) – c2t), w = 0.
(3.20) (3.21) (3.22)
It may now be shown that S lies on circle ACR. Circle EFO has u = (1/k)(b2c2(a4t(t – 1) + 2a2(t – 1)(b2(t – 1) – c2t) + b4(1 – t) + 2b2c2(t – 1) – c4t), v = (1/k)(a2c2t(a4t(t – 1) + b4(1 – t) + c4t), w = (1/k)(a2b2(1 – t)(a4t(t – 1) + 2a2(c2t – b2(t – 1)) + b4(t – 1) – 2b2c2 – c4t),
(3.23) (3.24) (3.25)
where k = 2(a2t – b2)(a2(t – 1) + c2)(b2(t – 1) – c2t).
(3.26)
It may now be shown that S lies on circle EFO. 4. The points A', B', C', D' and their properties 4
R'
W
To R'
To E'
E V F' D
D'
R
A U
P' S
S'
C'
A'
Q P
Q' O
F B
C B'
Fig. 2 Showing the cousin quadrilateral A'B'C'D' Circle WVA has u = 0, v = a2c2t(1 – t)(a2t – b2)/((a2t(t – 1) + 2b2(1 – t) + c2t)(a2(t – 1) – c2)), w = a2b2(1 – t)/(a2(t – 1) – c2).
(4.1) (4.2) (4.3)
This circle meets the circle ABCD at A and the point A' with co-ordinates A': x = a2t(b2 – a2t)(a2t(t – 1) + 2b2(1 – t) + c2t), y = b2(a2t + b2(t – 2) – c2t)(a2t(t – 1) + 2b2(1 – t) + c2t), z = (c2(a2t + b2(t – 2) – c2t))(t(b2 – a2t)).
(4.4) (4.5) (4.6)
It may now be checked that AA' passes through R, the midpoint of EF. 5
Points B', C', D' may also be obtained as the intersections of circles WVB, WVC, WVD with the circle ABCD. We call the cyclic quadrilateral A'B'C'D' the cousin of ABCD and CABRI II plus shows that it has the following properties (see Fig. 2): (i) A'C'^B'D' = U; (ii) A'B'^C'D' = E', a point on EF; (iii) D'A'^B'C' = F', a point on EF; (iv) The midpoint of A'C' = P' lies on circle UPQOS; (v) The midpoint of B'D' = Q' lies on circle UPQOS.
Flat 4, 12-14, Apsley Road, BRISTOL BS8 2SP.
6
Article: CJB/2011/202
The Midpoint Theorem Christopher Bradley Abstract: Let ABCD be a quadrangle, with L, M, N the midpoints of AB, BC, CA and U, V, W the midpoints of AD, BD, CD, then the conics AULMN, BVLMN, CWLMN have a fourth point in common. (D must not lie on the medians of ABC.)
D U A V W N
L
M
B P
Fig. 1 Three conics that have four points in common 1. The points L, M, N, U, V, W
1
C
We use areal co-ordinates throughout this short article with ABC as triangle of reference and D the point D(f, g, h). In order that none of the conics mentioned should degenerate then D must not lie on any of the medians of triangle ABC. The midpoints in this article therefore have co-ordinates: L: (0, 1. 1), M: (1, 0, 1), N: (1, 1, 0), U((1 + f)/2, g/2, h/2), V(f/2, (1 + g)/2, h/2), W(f/2, g/2, (1 + h)/2).
(1.1) (1.2)
2. The conics AULMN, BVLMN, CWLMN In areal co-ordinates conics have an equation of the form ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0.
(2.1)
The conic AULMN is found to have u = 0, v = – 2h((f + g – h + 1), w = 2g(f – g + h + 1), l = (f – g – h + 1)(h – g), m = – g(f – g + h + 1), n = h(f + g – h + 1).
(2.2)
The conic BVLMN is found to have u = 2h(f + g – h + 1), v = 0, w = 2f(f – g – h – 1), l = – f(f – g – h – 1), m = (f – g + h – 1)(h – f), n = – h(f + g – h + 1).
(2.3)
The conic CWLMN is found to have u = 2g(f – g + h + 1), v = 2f(f – g – h – 1), w = 0, l = – f(f – g – h – 1), m = – g(f – g + h + 1), n = (f + g – h – 1)(g – f).
(2.4)
3. The fourth common point P The three conics obviously have common points L, M, N. It may be shown now that they have a fourth point P in common with co-ordinates x = f(– f + g + h + 1)(– f2 + f + g2 – g(2h + 1) + h2 – h), (3.1) 2 2 2 y = g(f – g + h + 1)(f – f(2h + 1) – g + g + h – h), (3.2) 2 2 2 z = h(– f – g + h – 1)(– f + f(2g + 1) – g + g + h – h). (3.3) It can be checked that when D is the orthocentre of triangle ABC then P is the circumcentre of triangle ABC. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
Article: CJB/2011/203
Another Circle with Centre on the Brocard axis Christopher Bradley Abstract: In triangle ABC with symmedian point K, circumcentre O, circles BKC, CKA, AKB meet the sides at six further points, two on each side. It is shown these points lie on a circle with centre at a point O' on the Brocard axis such that OK = 2KO'.
F
E' G A
C' B' K O' F'
B
O
C
D
D' E
G' A'
1
1. The six points D, D', E, E', F, F' Throughout this article we use areal co-ordinates with ABC as triangle of reference. In areals the equation of a circle is of the form a2yz + b2zx + c2xy + (lx + my + nz)(x + y + z) = 0, (1.1) where l, m, n are to be determined from the co-ordinates of three points on the circle. For the circle BKC with K(a2, b2, c2), the symmedian point, we find m = 0, n = 0, and l = – 3b2c2/(a2 + b2 + c2).
(1.2)
This circle meets CA at the point E with co-ordinates E(a2 + b2 – 2c2, 0, 3c2) and it meets AB at the point F' with co-ordinates F'(c2 + a2 – b2, 3b2, 0). The other four points have co-ordinates that may now be found by cyclic change of x, y, z and a, b, c. To be precise these sets of co-ordinates are F: (3a2, b2 + c2 – 2a2, 0), D': (0, a2 + b2 – 2c2, 3c2), D: (0, 3b2, c2 + a2 – 2b2), E': (3a2, 0, b2 + c2 – 2a2). 2. Circle DEFD'E'F' The circle passing through D, E, F has an equation of the form (1.1) with u = 3b2c2(2a2 – b2 – c2)/(a2 + b2 + c2)2, v = 3c2a2(2b2 – c2 – a2)/(a2 + b2 + c2)2, w = 3a2b2(2c2 – a2 – b2)/(a2 + b2 + c2)2.
(2.1)
It may now be checked that points D', E', F' lie on this circle so all six points lie on a circle. The centre O' of this circle may now be obtained and has co-ordinates (x, y, z), where x = a2(3a2(b2 + c2) + 2b2c2 – a4 – 2b4 – 2c4), y = b2(3b2(c2 + a2) + 2c2a2 – b4 – 2c4 – 2a4), z = c2(3c2(a2 + b2) + 2a2b2 – c4 – 2a4 – 2b4). The Brocard axis OK is known to have equation b2c2(b2 – c2)x + c2a2(c2 – a2)y + a2b2(a2 – b2)z = 0.
(2.2)
(2.3)
It may now be checked that the point O' lies on the Brocard axis OK and is such that OK = 2KO'. This last fact was something of a surprise. Circle DEFD'E'F' and the circumcircle are two members of a coaxal system with radical axis the tangent to the Brocard circle at K. See the Figure. 2
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/204
Between the Triplicate Ratio Circle and the Circumcircle Christopher Bradley Abstract: When UVW is a triangle homothetic (and similar) to triangle ABC through K then the six points where UV, VW, WU intersect the sides of ABC lie on a circle. When UVW becomes a point at K one gets the Triplicate Ratio Circle and when U, V, W reach A, B, C one gets the Circumcircle. The centres O' of these circles move along the Brocard axis.
C'
A F E' U
B' K
O F'
O' E W
V
B
D'
D
A'
Fig. 1 Generating circles from triangle UVW
1
C
1. Introduction In this article we use areals throughout with ABC as triangle of reference. Point K is the symmedian point with co-ordinates (a2, b2, c2). For simplicity of working we take U, V, W to be the mid- points of AK, BK, CK respectively, but the main result is true whenever AU/UK = BV/VK = CW/WK. This main result is that the six points where UV, VW, WU meet the sides of ABC lie on a circle with centre on the Brocard axis OK, where O is the circumcentre of ABC. 2. The six points D, E, F, D', E', F' The midpoint U of AK has co-ordinates (x, y, z), where x = (2a2 + b2 + c2)/(2(a2 + b2 + c2)), y = b2/(2(a2 + b2 + c2)), z = c2/(2(a2 + b2 + c2)),
(2.1)
The co-ordinates of V, W may be obtained from (2.1) by cyclic change of both x, y, z and a, b, c. The equation of UV is c2(x + y) = (2a2 + 2b2 + c2)z.
(2.2)
The equations of VW and WU may be obtained from (2.2) again by cyclic change of both x, y, z and a, b, c. The line UV meets BC at D (0, 2a2 + 2b2 + c2, c2) and UV meets CA at E'(2a2 + 2b2 + c2, 0, c2). Similarly point E has co-ordinates (a2, 0, 2b2 + 2c2 + a2) and point F' has co-ordinates (a2, 2b2 + 2c2 + a2, 0). And finally F has co-ordinates (2c2 + 2a2 + b2, b2, 0) and D' has coordinates (0, b2, 2c2 + 2a2 + b2). 3. The circle DEFD'E'F' It may now be shown that the six points D, E, F, D', E', F' lie on the circle a2yz + b2zx + c2xy + (x + y + z)(lx + my + nz) = 0,
(3.1)
where l = (1/(4(a2 + b2 + c2)2)(b2c2(a2 + 2b2 + 2c2)) , and m, n follow from (3.2) by cyclic change of a, b, c. The centre O' of this circle has x-co-ordinate the complicated expression x = – 2a2(2a10 + 5a8(b2 + c2) + 2a6(b4 + b2c2 + c4) – 2a4(2b6 + 9b4c2 + 9b2c4 + 2c6) 2
(3.2)
– 2a2(2b8 + 11b6c2 + 18b4c4 + 11b2c6 + 2c8) – b10 – c2(7b8 + 16b6c2 + 16b4c4 + 7b2c6 + c8)). (3.3) The y- and z- co-ordinates follow from (3.3) by cyclic change of a, b, c. It may now be shown that O' lies on the Brocard axis OK with equation b2c2(b2 – c2)x + c2a2(c2 – a2)y + a2b2(a2 – b2)z = 0.
(3.4)
As U, V, W move along AK, BK, CK in harmony so O' moves along OK, being at O when U, V, W tend to A, B, C and at the mid point of OK when U, V, W all tend to K. 4. The points A', B', C' A further property is worth mentioning and that is that DF', ED' and AK concur at a point A' and similarly for points B' and C'. Point A' has co-ordinates (x, y, z), where x = – a2(2a2 + b2 + c2), y = b2(a2 + 2b2 + 2c2), z = c2(a2 + 2b2 + 2c2),
(4.1)
and where B', C' have co-ordinates which may be written down from (4.1) by cyclic change of x, y, z and a, b, c. It follows that triangles ABC and A'B'C' are in perspective with vertex K.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/2011/205
The Brocard Lines KΩ and KΩ' Christopher Bradley Abstract: Circles AΩΩ', BΩΩ', CΩΩ' meet the sides of ABC also in six further points which lie on two lines KΩ, KΩ' (which we call the Brocard lines). These six points also lie on four circles, each of the six points lying on two of the circles.
F' A
E' F
W K O
B
W'
D'
D
C
E
Fig. 1 More circles with centres on the Brocard axis 1. Circles AΩΩ', BΩΩ', CΩΩ' We use areal co-ordinates with ABC triangle of reference. Point Ω has co-ordinates (1/b2, 1/c2, 1/a2) and point Ω' has co-ordinates (1/c2, 1/a2, 1/b2). The equation of the circle AΩΩ' is now found to be of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0, (1.1) where u = 0, v = c2a2(b2 – a2)/(a4 – b2c2), w = a2b2(c2 – a2)/(a4 – b2c2). 1
(1.2)
The equations of circles BΩΩ' and CΩΩ' can be written down from (1.2) by cyclic change of x, y, z and a, b, c. 2. Points D, D', E, E', F, F' Circle AΩΩ' meets AB at the point F' with co-ordinates (a2(a2 – b2), b2(a2 – c2), 0) and it meets CA at E with co-ordinates (a2(a2 – c2), 0, c2(a2 – b2)). Circle BΩΩ' meets BC at D' with coordinates (0, b2(b2 – c2), c2(b2 – a2)) and it meets AB at F with co-ordinates (a2(b2 – c2), b2(b2 – a2), 0). Circle CΩΩ' meets CA at E' with co-ordinates (a2(c2 – b2), 0, c2(c2 – a2) and it meets BC at D with co-ordinates (0, b2(c2 – a2), c2(c2 – b2)). 3. Lines DEFKΩ and D'E'F'KΩ' It may now be shown that D, E, F, K, Ω lie on the line with equation x/(a2(b2 – c2)) + y/(b2(a2 – b2)) + z(c2 – a2)/(c2(b2 – c2)(a2 – b2)) = 0.
(3.1)
And also it may be shown that D', E', F', K, Ω' lie on the line with equation x/(a2(a2 – b2)) + y/(b2(c2 – a2)) + z(b2 – c2)/(c2(c2 – a2)(a2 – b2)) = 0.
(3.2)
A pause here is important to think why these equations do not show symmetry. 4. Circles DD'E'F, EE'F'D, FF'D'E It may further be checked that points D, D', E', F lie on a circle with an equation of the form (1.1) with u = (1/k)(b2c2(a2 – b2)(c2 – a2)(b2 + c2)), (4.1) 2 4 2 2 2 2 v = (1/k)(a c (a – b )(c – b )), (4.2) 2 4 2 2 2 2 w = (1/k)(a b (a – c )(b – c )), (4.3) where k = (a2b2 – c4)(a2c4 – b4).
(4.4)
Equations of circles EE'F'D, FF'D'E follow by cyclic change of u, v, w and a, b, c.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
3
Article: CJB/2011/206
Dividing OH into Five or Seven Equal Parts Christopher Bradley Abstract: By dividing each side of a triangle into quarters and then drawing the resulting nine Cevian lines, six points are created internally through which two circles are drawn. Their centres lie on the Euler line and divide OH in the ratios 1:4 and 4:3.
A
L3
N2 U
M3 R
M2
Q X
O
G N3
S H L2
V
W P
B
L1
M1
N1
C
Fig. 1 Showing the centres X and S on the Euler line 1. The Nine Starting Points on the Sides of ABC and the resulting Nine Cevian Lines Areal co-ordinates are used throughout with ABC as triangle of reference. The co-ordinates of the nine points are as follows: L1: (0, 3, 1), L2(1, 0, 3), L3(3, 1, 0), M1(0, 1, 1), M2(1, 0, 1), M3(1, 1, 0), 1
N1(0, 1, 3), N2(3, 0, 1), N3(1, 3, 0).
(1.1)
Cevians have equations as follows: AL1: y = 3z, BL2: z = 3x, CL3: x = 3y, AM1: y = z, BM2: z = x, CM3: x = y. AN1: z = 3y, BN2: x = 3z, CN3: y = 3x.
(1.2)
2. Points U, V, W, P, Q, R U = AM1^BN2^CL3 has co-ordinates (3, 1, 1), V = AL1^BM2^CN3 has co-ordinates (1, 3, 1), W = AN1, BL2, CM3 has co-ordinates (1, 1, 3). (2.1) P= AM1^BL2^CN3 has co-ordinates (1, 3, 3), Q = AN1^BM2^CL3 has co-ordinates (3, 1, 3) R = AL1^BN2^CM3 has co-ordinates (3, 3, 1). (2.2) 3. Circle UVW and its centre X In areal co-ordinates circles have equations of the form a2yz + b2zx + c2xy + (x + y + z)(lx + my + nz) = 0,
(3.1)
where the constants l, m, n are calculated from the co-ordinates of three point through which the circle passes. Circle UVW has l = (1/25)(a2 – 4b2 – 4c2), m = (1/25)(b2 – 4c2 – 4a2), n = (1/25)(c2 – 4a2 – 4b2).
(3.2)
When expressed without fractions in the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0
(3.3)
the circle has u = 4(b2 + c2) – a2, v = 4(c2 + a2) – b2, w = 4(a2 + b2) – c2, f = (1/2)(3(b2 + c2) – 17a2), g = (1/2)(3(c2 + a2) – 17b2), h = (1/2)(3(a2 + b2) – 17c2).
(3.4)
In terms of these constants the centre X has x-co-ordinate x = vw – gv – hw – f2 + fg + hf.
(3.5)
which gives x = ( – 125/4)(3a4 + b4 + c4 – 2b2c2 – 4a2(b2 + c2)) The y- and z- co-ordinates of X follow from (3.6) by cyclic change of a, b, c.
2
(3.6)
It may now be checked that X lies on OH and is such that XH = 4OX, so that X is 1/5 th of the way from O to H. 4. Circle PQR and its centre S Circle PQR has an equation of the form (3.1) with l = (3/49)(3a2 – 4(b2 + c2)), m = (3/49)(3b2 – 4(c2 + a2)), n = (3/49)(3c2 – 4(a2 + b2)).
(4.1)
When re-expressed in the form of (3.3) we find u = 3(3a2 – 4b2 – 4c2), v = 3(3b2 – 4c2 – 4a2), w = 3(3c2 – 4a2 – 4b2) f = (1/2)(25a2 – 3(b2 + c2)), g = (1/2)(25b2 – 3(c2 + a2)), h = (1/2)(25c2 – 3(a2 + b2)).
(4.2)
Using (3.5) we find the x-co-ordinate of the centre S to be x = – (343/4)(a4 + 3(b4 + c4) – 6b2c2 – 4a2(b2 + c2)).
(4.3)
The y- and z- co-ordinates of S follow from (4.3) by cyclic change of a, b, c. It may now be checked that S lies on OH and is such that 4OS = 3SH, so that S is 3/7th of the way from O to H. It may be remarked that by choosing different positions for the L- and N- points on the sides of the triangle then provided there is symmetry the centres of the circles formed all lie on OH with positions dividing the line OH in different fractions.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/2011/207
The Median Conic Christopher Bradley Abstract: Given a triangle ABC and its centroid G, the midpoints of AG, BG, CG are denoted by U, V, W. Conics BCUVW, CAUVW, ABUVW are drawn and meet the sides again at point D, D', E, E', F, F'. It is proved that D, D', E, E', F, F' lie on a conic.
A
F
E' U
F'
G V
B
D
E W
D'
Fig. 1 Three conics producing a fourth
1. The three conics
C
We use areal co-ordinates throughout with ABC as triangle of reference. So the centroid G has co-ordinates G (1/3, 1/3, 1/3). It follows that U, V, W the midpoints of AG, BG, CG respectively have co-ordinates U(2/3, 1/6, 1/6), V(1/6, 2/3, 1/6) and W(1/6, 1/6, 2/3). Conics in areal co-ordinates have equations of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(1.1)
where the constants u, v, w, f, g, h are determined from the five points lying on the conic. It follows that for conic BCUVW u = 6, v = 0, w = 0, f = 8, g = – 7, h = – 7
(1.2)
The conics CAUVW, ABUVW follow from Equation (1.2) by cyclic change of u, v, w and f, g, h. 2. The points D, E, F, D', E', F' and the conic on which they lie The conic BCUVW meets CA at the point E' with co-ordinates (7, 0, 3). Similarly E has coordinates (3, 0, 7). Likewise D has co-ordinates (0, 7, 3) and D' has co-ordinates (0, 3, 7) and again F has co-ordinates (7, 3, 0) and F' has co-ordinates (3, 7, 0). It may now be shown that these six points lie on the conic with equation 21(x2 + y2 + z2) – 58(yz + zx + xy) = 0.
(2.1)
Two further points may be made as shown by CABRI software: (1) G may be replaced by any other point on the three medians and one still obtains a conic, (2) No other point in the plane of ABC (not on the sides) has the property of defining a conic.
Flat 4, Terrill Court, 12- 14, Apsley Road, Clifton, BRISTOL BS8 2SP.
Article: CJB/2011/208
A Pair of Isogonal Conjugates produce a Conic through 6 Points Christopher Bradley Abstract: If P and Q are an isogonal conjugate pair then circles APQ, BPQ, CPQ intersect the sides of triangle ABC in six points and it is found that a conic passes through them. The construction only produces a conic when P and Q are isogonal conjugates. When P and Q are the Brocard points then the conic degenerates into a pair of lines.
A
Oa F' X Y
P
E Q
W
F
C D
B
D'
Ob E' Oc
Fig. 1
1
1. Points P and Q and Circle APQ We use areal co-ordinates throughout this article. We label P by the co-ordinates (f, g, h) and Q, its isogonal conjugate, by (a2/f, b2/g, c2/h). In areals a circle has an equation of the form a2yz + b2zx + c2xy + (x + y + z)(lx + my + nz) = 0,
(1.1)
where l, m, n are to be found by using the co-ordinates of the three given points lying on the circle. We find that the circle APQ has l = 0, m = (1/k)(c2g(a4g2h2 + a2h(2c2fg2 – b2h(f2 + 2hf + g2 + 2gh + h2)) + f2(b2h + c2g)2), n = – (1/k)(b2h(a4g2h2 + a2g(2b2fh2 – c2g(f2 + 2fg + g2 + 2gh + h2)) + f2(b2h + c2g)2), and k = (f + g + h)(a2gh + f(b2h + c2g))(b2h2 – c2g2).
(1.2)
(1.3)
2. The six points D, D', E, E', F, F' and the conic through these points Circle APQ meets CA at E with co-ordinates x = h(a4g2h2 + a2g(2b2fh2 – c2g(f2 + 2fg + g2 + 2gh + h2)) + f2(b2h + c2g)2, y = 0, z = (a2gh2 + f(b2h2 – c2g(f + g)))((g + h)(b2h + c2g) – a2gh) Circle APQ meets AB at F' with co-ordinates x = g(a4g2h2 + a2h(2c2fg2 – b2h(f2 + 2hf + g2 + 2gh + h2)) + f2(b2h + c2g)2), y = (a2g2h – f(b2h(f + h) – c2g2))((g + h)(b2h + c2g) – a2gh), z = 0.
(2.1)
(2.2)
The co-ordinates of F, D', D, E' follow from Equations (2.1) and (2.2) by cyclic changes of x, y, z and a, b, c and f, g, h. (F and D from (2.1) and D' and E' from (2.2)). It may now be shown that these six points lie on a conic. Unfortunately its equation took up to three pages of DERIVE output and so is impossible to copy down without error. I regret it must be left to the reader. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
Article: CJB/209/2012
Three Concentric Circles Christopher Bradley bd
F
ab A da
P
D
A' B
Q
S B'
D'
E O
cd
ac
C' R C
bc
Fig. 1 Nine circles including five coaxal pairs Abstract: Starting with a cyclic quadrilateral ABCD with centre O, the midpoints A', B', C', D' of OA, OB, OC, OD respectively form a second cyclic quadrilateral with half the size. It is shown that points B, B', C', C lie on a circle and so do C, C', D', D etc. The centre of BB'C'C is labelled bc, and that of CC'D'D is labelled cd etc. Five pairs of these circles are immediately seen to be coaxal and analysis is given for one such pair. 1
1. The Points We take ABCD to have equation x2 + y2 = 1, with O(0, 0) and give points co-ordinates in the usual manner with A((1 – a2)/(1 + a2), 2a/(1 + a2)) and similarly with B, C, D having parameters b, c, d. Point A' being halfway from A to O has co-ordinates half those of A, B, C, D and circle A'B'C'D' has equation x2 + y2 = ¼. 2. Circles and their centres It is easy to show that circle BB'C'C has equation x2 + y2 – 3(1 – bc)/(2(1 + bc))x – 3(b + c)/(2(1 + bc))y + ½ = 0.
(2.1)
Its centre bc has co-ordinates (x, y) where x = (3/4)(1 – bc)/(1 + bc), y = (3/4)bc(b + c)/(1 + bc).
(2.2)
There are 5 more such circles CC'D'D, DD'A'A, AA'B'B, AA'C'C and BB'D'D with centres cd, da, ab, ac, bd respectively with equations and co-ordinates duly adjusted with change of parameters. 3. Lines of interest It may now be shown that centres ab ac da lie on a line with equation y = x(a2 – 1)/2a + 3(a2 + 1)/(8a).
(3.1)
Similarly ab, bc, bd lie on a line as do ac, bc, cd and as do da, bd, cd with the same equation as (3.1) but with parameters b, c, d replacing a. Line OA has equation y = 2ax/(1 – a2),
(3.2)
which can be seen to be perpendicular to ab ac da. Similarly OB, OC, OD are perpendicular respectively to lines ab bc bd, ac bc cd and da bd cd. Circles centre ac and bd intersect at two points E and F and EF has slope 2(bd – ac)/(abc – abd + acd – bcd – a + b – c + d),
(3.3)
which of course is perpendicular to the line ac bd. In this way the pairs of coaxal circle arise.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP 2
Article: 210/CJB/2012
12 Points, 8 six – point Conics, 4 Conics through each Point Christopher Bradley
Line 1
Line 2
Line 3
A 23 15
35
25
13
45
Line 4
O N 36
Line 5
H 16
24 C
46
B 14
26
Line 6
Abstract: Given a triangle ABC from the midpoints of each side perpendiculars are drawn to the other two sides. These lines are labelled Lines 1 to 6 as in the Figure. Their intersections create 12 finite points, so that, for example 35 is the intersection of Lines 3 and 5. It is found that 8 sixpoint conics can be drawn through these 12 points, with 4 of the conics through each point. The nine-point circle is, of course, one of the conics. In a later article some properties of these conics are established. 1. The Lines 1 to 6 Line 1 is the line parallel to AH passing through the midpoint of AB and has equation 1
(c2 + a2 – b2)(x – y) + (3a2 + b2 – c2)z = 0.
(1.1)
Line 3 is the line parallel to BH passing through the midpoint of AB and has equation (a2 – b2 – c2)(x – y) + (a2 + 3b2 – c2)z = 0.
(1.2)
Line 4 is the line parallel to BH passing through the midpoint of BC and has equation (3b2 + c2 – a2)x + (a2 + b2 – c2)(y – z) = 0.
(1.3)
Line 6 is the line parallel to CH passing through the midpoint of BC and has equation (b2 + 3c2 – a2)x + (b2 – c2 – a2)(y – z) = 0.
(1.4)
Line 5 is the line parallel to CH passing through the midpoint of CA and has equation (b2 + c2 – a2)(z – x) + (a2 – b2 + 3c2)y = 0.
(1.5)
Line 2 is the line parallel to AH passing through the midpoint of CA and has equation (c2 – a2 – b2)(z – x) + (c2 + 3a2 – b2)y = 0.
(1.6)
2. The 12 points There are three midpoints with co-ordinates 46: (0, ½, ½), 25: (½, 0, ½), 13: (½, ½, 0).
(2.1)
Lines 1 and 4 intersect at 14 with co-ordinates x = – (a2 + b2 – c2)2, y = – a4 + 2a2(2b2 + c2) + (b4 – c4), z = 2b2(c2 + a2 – b2).
(2.2)
Lines 1 and 5 intersect at 15 with co-ordinates x = – a4 – 4a2c2 + (b2 – c2)2, y = 2a2(a2 – b2 – c2), z = (c2 + a2 – b2)2 .
(2.3)
Lines 1 and 6 intersect at 16 with co-ordinates x = – (a2 + b2 – c2)(c2 + a2 – b2), y = 2(a4 – a2(b2 + 2c2) – c2(b2 – c2)), z = (a2 – b2 – c2)(a2 – b2 + c2).
(2.4)
Lines 2 and 3 meet at 23 with co-ordinates x = – a4 – 4a2b2 – (b2 – c2)2, y = (a2 + b2 – c2)2, z = 2a2(a2 – b2 – c2) .
(2.5)
Lines 2 and 4 meet at 24 with co-ordinates x = (b2 – c2)2 – a4, y = (a2 + b2 – c2)(a2 – b2 – c2), z = 2(a4 – a2(2b2 + c2) + b2(b2 – c2)). (2.6) Lines 2 and 6 meet at 26 with co-ordinates x = (c2 + a2 – b2)2, y = – 2c2(a2 + b2 – c2), z = a4 – 2a2(b2 + 2c2) + (b4 – c4). 2
(2.7)
Lines 3 and 5 meet at 35 with co-ordinates x = 2((b2 – c2)2 – a2(b2 + c2)), y = (a2 + b2 – c2)(a2 – b2 – c2), z = (a2 – b2 – c2)(a2 – b2 + c2),
(2.8)
Lines 3 and 6 meet at 36 with co-ordinates x = – 2b2(c2 + a2 – b2), y = a4 – 2a2c2 – b4 – 4b2c2+ c4, z = (b2 + c2 – a2)2.
(2.9)
Lines 4 an 5 meet at 45 with co-ordinates x = 2c2(a2 + b2 – c2), y = – (b2 + c2 – a2)2, z = – (a4 – 2a2b2 + b4 – 4b2c2 – c4).
(2.10)
Note that points 35, 16 and 24 lie at the midpoints respectively of AH, BH and CH. There are other symmetries in the Figure. For example N is the midpoint of 15 26, 14 23, 36 45. 3. The 8 Conics Cabri II plus shows that the following sets of 6 points each lie on a conic: (1) (2) (3) (4) (5) (6) (7) (8)
13, 16, 24, 25, 35, 46 (the nine-point circle), 14, 15, 23, 26, 36, 45, 14, 16, 23, 25, 36, 45, 14, 15, 23, 26, 35, 46, 13, 15, 24, 26, 35, 46, 14, 16, 23, 25, 35, 46, 13, 15, 24, 26, 36, 45, 13, 16, 24, 25, 36, 45.
Note that each of the 12 points lies on 4 of these conics. The algebra obtaining the equations of these conics is very involved, so we only provide the analysis for Conic (2), the equation of the nine-point circle being well-known. Equations of the other conics have been obtained but are not supplied in this article. Conic (2) has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, where we find u = – a8 + 4a6(b2 + c2) – 2a4(2b4 + 7b2c2 + 2c4) + 8a2b2c2(b2 + c2) + b8 + 2b6c2 – 6b4c4 + 2b2c6 + c8, v = a8 + 2a6c2 – 2a4(2b4 – 4b2c2 + 3c4) + 2a2(b2 – c2)(2b4 – 5b2c2 – c4) + (c2 – b2)(b6 – 3b4c2 + b2c4 + c6),
3
(3.1)
(3.2) (3.3)
w = a8 + 2a6b2 – 2a4(3b4 – 4b2c2 + 2c4) + 2a2(b2 – c2)(b4 + 5b2c2 – 2c4) + (b2 – c2)(b6 + b4c2 – 3b2c4 + c6), f = – a8 – a6(b2 + c2) + a4(5b4 – 12b2c2 + 5c4) – 3a2(b6 – b4c2 – b2c4 + c6) + 2b2c2(b2 – c2)2, g = a6(2c2 – 3b2) + a4(5b4 + 3b2c2 – 4c4) – a2(b6 + 12b4c2 – 3b2c4 – 2c6) – b2(b6 + b4c2 – 5b2c2 + 3c6), h = a6(2b2 – 3c2) + a4(5c4 + 3b2c2 – 4b4) – a2(c6 + 12c4b2 – 3c2b4 – 2b6) – c2(c6 + c4b2 – 5c2b2 + 3b6).
Flat 4, Terrill Court, 12 – 14, Apsley Road , BRISTOL BS8 2SP.
4
(3.4) (3.5) (3.6) (3.7)
A CYCLIC QUADRILATERAL AND ITS MIDPOINT CIRCLES CHRISTOPHER BRADLEY
A S L
D F T R
O
E X
P
N C G
M Q
B
FIGURE 1
ABSTRACT: A CYCLIC QUADRILATERAL ITS MIDPOINTS AND THE SIX MIDPOINT CIRCLES
1
1. Introduction Let ABCD be a cyclic quadrilateral, P, Q, R, S the midpoints of the sides AB, BC, CD, DA respectively, O the centre of circle ABCD and E the intersection of the diagonals AC and BD. It is shown that APOS, BQOP, CROQ, DSOR are circles, having the same radius and their centres L, M, N, T are concyclic. Suppose that circle APOS meets circle CROQ at F and circle BQOP meets circle DSOR at G, then it is shown that O, E, F, G are concyclic. If the centre of circle OEFG is denoted by X then it is shown that MXT is a line parallel to BD and NXL is a line parallel to AC. 2. The points P, Q, R, S and the circles APOS, BQOP, CROQ, DSOR We take the circle ABCD to have equation x2 + y2 = 1 and A to have co-ordinates (x, y),where x = (1 – a2)/(1 + a2) and y = 2a/(1 + a2). (2.1) Points B, C, D have similar co-ordinates but with b, c, d instead of a. The midpoints of AO, BO, CO, DO are denoted by L, M, N, T respectively so that they have co-ordinates that are ½ of those of A, B, C, D. The equation of the circle APOS centre L passing through O and A is (1 + a2)(x2 + y2) – (1 – a2)x – 2ay = 0,
(2.2)
and this meets AB with equation (1 – ab)x + (a + b)y = (1 + ab),
(2.3)
at the point P with co-ordinates (x, y), where x = (1 – a2b2)/((1 + a2)(1 + b2)), y = (a + b)(1 + ab)/((1 + a2)(1 + b2)).
(2.4)
Circles BQOP, CROQ, DSOR and points Q, R, S have similar equations an co-ordinates with appropriate changes of parameters. 3. Points E, F, G, the circle OEFG and its centre X Circles APOS and CROQ meet at the point F with co-ordinates (x, y), where x = (1 – a2c2/((1 + a2)(1 + c2)), y = (1 + ac)(a + c)/((1 + a2)(1 + c2)).
(3.1)
Circles BQOP and DSOR meet at the point G with co-ordinates (x, y), where x = (1 – b2d2)/((1 + b2)(1 + d2)), y = (1 + bd)(b + d)/((1 + b2)(1 + d2)).
(3.2)
2
We can now obtain the equation of the circle OFG which is (abc + acd – abd – bcd + a – b + c – d)(x2 + y2) + (abc – abd + acd – bcd – a + b – c + d)x – 2(ac – bd)y = 0. (3.3) And it may be checked that this circle also passes through E = AC^BD with co-ordinates (x, y), where x = – (1/k)(abc – abd + acd – bcd – a + b – c + d), y = (1/k)(2(ac – bd)) and k = (abc – abd + acd – bcd + a – b + c – d). (3.3) The centre X of this circle therefore has co-ordinates (x, y), where x = – (1/(2k))(abc – abd + acd – bcd – a + b – c + d), y = –(1/k)(bd – ac), k = (abc – abd + acd – bcd + a – b + c – d).
(3,4)
It may now be checked that the line MT passes though X and is parallel to BC and also that NL passes through X and is parallel to AC.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
.
3
Article: cjb/212/2012
Quarter size Circles at Triangle Vertices Christopher Bradley Abstract: In triangle ABC points Q, R lie on side BC so that Q is a quarter of the way from B to C and R is a quarter of the way from C to B. Points S, T on CA and points U, V on side AB are similarly positioned. We prove circles AUT, BQV, CTR have equal radius.
A
U
V
B
T
O
S
Q R
Figure The three corner circles
1
C
1. The co-ordinates of points The final result is obvious from pure geometry since the three circles are reductions of the circumcircles at B, C, A by a factor of four. However we give as usual a co-ordinate proof. We take the circumcircle to be the unit circle and then B has co-ordinates x = (1 – b2/(1 + b2), y = 2b/(1 + b2).
(1.1)
The co-ordinates of C, A follow at once, by replacing b with c, a respectively. The co-ordinates of Q follow immediately from those of B and C and are x = 3/(2(1 + b2)) + 1/(2(1 + c2)) – 1, y = 6b/(4(1 + b2)) + 3c/(4(1 + c2)).
(1.2) (1.3)
Co-ordinates of R, S, T, U, V follow at once by appropriate changer of parameters. 2. Circle BQV and its radius Circle BQV has an equation of the form x2 + y2 + 2gx + 2fy + k = 0.
(2.1)
By inserting it turn the co-ordinates of B, Q and V we obtain three equations whose solution for g, f, k are g = 3(b2 – 1)/(4(b2 + 1)), f = – 3b/(2(b2 + 1)), k = ½. (2.2) The radius squared is given by g2 + f2 – k = 1/16. So the radius is ¼ that of the circumcircle and as this is independent of a b, c it is also the radius of circle CSR and AUT.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/2012/213
When an Incircle produces a Circumcircle Christopher Bradley Abstract: A cyclic quadrilateral ABCD with AC perpendicular to BD is such that the tangents at A, B, C and D form a cyclic quadrilateral. This configuration is described analytically. If AC^BD = X and the centres of the two circles are O and Y, then it is proved that YOX is a line.
G E
To G To F
W A
To E D
S
V X
T
R
P O
Y
C Q
B
U
Figure The two cyclic quadrilaterals ABCD and TUVW
1
1. The circle ABCD We take ABC to be the triangle of reference with A(1, 0, 0), B(0, 1, 0), C(0, 0, 1). For D we start by taking any point on the circumcircle a2yz + b2zx + c2xy = 0 (1.1) with co-ordinates D(–a2t(1 – t), b2(1 – t), c2t), where t is a parameter. The co-ordinates of D must be normalized so that they must be divided by (–a2t(1 – t) + b2(1 – t) + c2t), and given labels. The displacements AC are denoted by (f, g, h) = (– 1, 0, 1) and the displacement BD is denoted by (u, v, w) The condition for BD and AC to be perpendicular is well-known to be a2(gw + hv) + b2(hu + fw) + c2(fv + gu) = 0. (1.2) This provides a value for the parameter t, which is t = (a4 – a2(b2 + 2c2) – c2(b2 – c2))/(a2(a2 – b2 – c2)). (1.3) The equations of the tangents at A, B, C, D may now be written down and they are A: c2y + b2z = 0, B: c2x + a2z = 0, C: b2x + a2y = 0, D: a2b2(b2 + c2 – a2)2x + (a4 – a2(b2 + 2xc2) – c2(b2 – c2)2y + b2c2(a2 + b2 – c2)2z.
(1.4) (1.5) (1.6) (1.7)
2. The circle TUVW The tangents at A and B meet at T with co-ordinates T(a2, b2, – c2), the tangents at B and C meet at U with co-ordinates U(– a2, b2, c2), the tangents at C and D meet at V with co-ordinates V(x, y, x), where x = – a2(a2 + b2 – c2), (2.1) 2 2 2 2 y = b (a + b – c ), (2.2) 4 2 2 2 2 2 2 z = 2a – a (2b + 3c ) – c (b – c ). (2.3) The tangents at D and A meet at W with co-ordinates W(x, y, z), where x = a4 – a2(b2 + 3c2) – 2c2(b2 – c2), y = b2(b2 + c2 – a2), z = – c2(b2 + c2 – a2). The equation of the circle TUVW may now be determined. It has the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0, where we find u = (1/k)(a2b2c2(a2 – b2 – c2))/(a2 + b2 – c2), v = a2b2c2/((a2 + b2 – c2)(a2 – b2 – c2)), 2
(2.4) (2.5) (2.6)
(2.7) (2.8) (2.9)
w = (1/k)(a2 b2c2(a2 + b2 – c2))/(a2 – b2 – c2),
(2.10)
k = (a4 – 2a2(b2 + c2) + (b2 – c2)2).
(2.11)
and
3. The points O, X, Y lying on a line The point O, the centre of circle ABCD has well-known co-ordinates (x, y, z) given by x = a2(b2 + c2 – a2), y = b2(c2 + a2 – b2), z = c2(a2 + b2 – c2).
(3.1) (3.2) (3.3)
The point X = AC^BD has co-ordinates X(x, y, z), where x = a2 + b2 – c2, y = 0, z = b2 + c2 – a2.
(3.4) (3.5) (3.6)
It may now be checked that X lies on both TV and UW. The centre Y of circle TUVW may be calculated by supposing it has co-ordinates Y(p, q, r), where p + q + r = 1 and TY2 = UY2 = VY2. This is a technically difficult matter which leads to the result p = 1 + (1/k)(a10(b2 + c2) – a8(3b4 + 6b2c2 + 5c4) + 2a6(b6 + 4b4c2 + 4b2c4) + + 2a4(b8 – b6c2 + b2c6 – 5c8) – a2(b6 – 3b4c2 + 3b2c4 – c6)(3b4 + 6b2c2 + 5c4) + + (b2 – c2)(b4 – c4)(b6 – 3b4c2 + 3b2c4 – c6)), (3.7) 2 10 8 2 2 6 2 2 2 4 6 4 2 2 4 6 q = – (1/k)b (a – 3a (b + c ) + 2a (b + c ) + 2a (b + b c – b c + c ) – 2 8 – a (3b – 2b4c4 – 4b2c6 + 3c8) + b10 – 3b8c2 + 2b6c4 + 2b4 c6 – 3b2c8 + c10), (3.8) 2 10 8 2 2 6 4 2 2 4 4 6 4 2 r = – (1/k)c (a – a (3b + 5c ) + 2a (2b + 3b c + 5c ) – 2a (2b – b c + 5c6) + + a2(3b8 + 2b6c2 – 4b4c4 – 6b2c6 + 5c8) – b10 + 3b8c2 – 2b6c4 + 3b2c8 – c10). (3.9) and k = (a2 + b2 – c2)(a2 – b2 – c2)(a4 – 2a2(b2 + c2) + (b2 – c2)2). It may now be checked that the points O, X, Y lie on the line with equation b2(c2 + a2 – b2)(b2 + c2 – a2)x – (a6 – a4(2b2 + 3c2) + a2(b4 + 3c4) – – c2(b2 – c2)2)y – b2(a2 + b2 – c2)(c2 + a2 – b2)z = 0. 4. The polar of X with respect to circle ABCD
3
(3.10)
(3.11)
The equations of the sides of cyclic quadrilateral TUVW are calculated to be TU: c2x + a2z = 0. UV: b2x + a2y = 0. VW: a2b2(b2 + c2 – a2)2x + (a8 – 2a6(b2 + 2c2) + a4(b4 + 2b2c2 + 6c4) + + 2a2c2(b2 – c2)(b2 + 2c2) + c4(b2 – c2)2)y + b2c2(a2 + b2 – c2)2z = 0. WT: c2y + b2z = 0.
(4.1) (4.2) (4.3) (4.4)
These are, of course, the tangents at B, C, D, A respectively (see Section 1). UV and WT meet at a point E with co-ordinates E(a2, – b2, c2). TU and VW meet at F with co-ordinates (x, y, z), where x = – a2(a4 – a2(b2 + 2c2) - b2c2 + c4), (4.5) 2 4 2 2 2 2 4 y = b (a – a b + b c – c ), (4.6) 2 4 2 2 2 2 2 4 z = c (a – a (b + 2c ) - b c + c ). (4.7) The equation of EF may now be worked out and is b2(a2 – b2 – c2)x + (a4 – a2(b2 + 2c2) – c2(b2 – c2))y – b2(a2 + b2 – c2)z = 0. (4.8) It may now be checked that this is the polar of X with respect to circle ABCD. 5. Known facts We conclude by stating a number of known facts true for all cyclic quadrilaterals ABCD whether AC and BD are at right angles or not. (1) (2) (3) (4)
A circle passes through E, A, O, C; A circles passes through F, B, O, D; These circles intersect at O and at a point G lying on EF. In this configuration G has co-ordinates (x, y, z), where x = – a2(a6 – a4(b2 + 3c2) + a2(3c4 – b4) + b6 – b4c2 + b2c4 – c6), y = b2(a2 + b2 – c2)(c2 + a2 – b2)(a2 – b2 – c2), z = c2(a2 – b2 – c2)(a4 + b4z + c4 – 2(a2 + b2)c2).
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(5.1) (5.2) (5.3)
Article: CJB/2012/214
The Super-Cevian Triangles, their Conic and Two Perspectives Christopher Bradley
Z
U A Y
F E P R Q V B
D W
X
1
C
Abstract: In the above Figure, ABC is a triangle, P is a Cevian point and D, E, F the feet of the Cevians on BC, CA, AB respectively. The triangle DEF is the Cevian triangle. A triangle XYZ is constructed by drawing through A a line parallel to EF and through B and C lines parallel to FD and DE respectively. Also a triangle UVW is constructed by drawing lines parallel to EF, FD, DE through D, E, F respectively. These two triangles we term as Super-Cevian triangles. We then prove that ABCUVW is a conic. Q is the perspector of triangles UVW and XYZ and R is the perspector of triangles ABC and XYZ. These results are established in this article, using areal co-ordinates with ABC as triangle of reference. 1. The triangles DEF and XYZ We take P to have co-ordinates P(l, m, n). The feet of the Cevians are then D(0, m, n), E(l, 0, n) and F(l, m, 0). The equations of the sides of triangle DEF may now be calculated and are EF: – mnx + nly + lmz = 0, (1.1) FD: mnx – nly + lmz = 0, (1.2) DE mnx + nly – lmz = 0. (1.3) The line through A parallel to EF has equation n(l + m)y + m(n + l)z = 0. Similarly the line through B parallel to FD has equation n(l + m)x + l(m + n)z = 0. And the line through C parallel to DE has equation m(n + l)x + l(m + n)y = 0.
(1.4) (1.5) (1.6)
The co-ordinates of X may be obtained as the intersection of the last two lines and are X(– l(m + n), m(n + l), n(l + m)). Similarly the co-ordinates of Y and Z are Y(l(m +n), – m(n + l), n(l + m)) and Z(l(m + n, m(n + l), – n(l + m)) respectively. R is the perspector of triangle ABC and XYZ and evidently has co-ordinates R(l(m + n), m(n + l), n(l + m)). 2. Triangle UVW The line parallel to EF through D has equation mn(2l + m + n)x + l(m – n)(ny – mz) = 0.
(2.1)
Similarly the lines parallel to FD and DE through the points E and F respectively have equations nl(2m + n + l)y + m(n – l)(lz – nx) = 0 (2.2) and lm(2n + l + m)z + n(l – m)(mx – ly) = 0. (2.3) 2
The co-ordinates of U may be obtained as the intersection of the last two lines and are U: (l(m + n)(2l + m + n), m(n + l)(n – l), n(m + l)(m – l)).
(2.4)
Those of V and W may now be written down by cyclic change of x, y, z and l, m, n and are V: (l(n + m)(n – m), m(n + l)(2m + n + l), n(l + m)(l – m)). (2.5) W: (l(m + n)(m – n), m(l + n)(l – n), n(l + m)(2n + l + m)). (2.6) Q is the perspector of triangles UVW and XYZ and has co-ordinates Q(l(m + n)2,m(n + l)2, n(l + m)2.
(2.7)
3. The conic ABCUVW Since this conic passes through A, B, C its equation has the form fyz + gzx + hxy = 0.
(3.1)
By substituting into Equation (3.1) the co-ordinates of U, V, W in turn we obtain three equations for f, g, h with solution f = l(m + n)(m2 – n2)(2l + m + n), g = m(n + l)(n2 – l2)(2m + n + l), h = n(l + m)(l2 – m2)(2n + l + m). It may now be checked that the point Q lies on this conic.
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(3.2) (3.3) (3.4)
Article: CJB/2012/215
Multiplication of Points using Barycentric Co-ordinates Christopher Bradley Abstract: If you have points with barycentric co-ordinates (f, g, h), (u, v, w) then the rule of multiplication is that the result has co-ordinates (fu, gv, hw). It is shown in this article how to perform this multiplication using a geometric construction . The method is as follows: First draw the circumconic fx + gy + hz = 0. Next take the isotomic conjugation of (u, v, w) to get the point (1/u, 1/v, 1/w). To conjugate this point finally use the circumconic to perform the second conjugation taking (1/u, 1/v, 1/w) to the product point (f/(1/u), g/(1/v), h/(1/v)) = (fu, gv, hw). This is evidently a commutative product and so the product may also be obtained using the conic ux + vy + wz = 0 and operating the two conjugations on the point (f, g, h).
A Brocard axis
Z
Y Y'
Z' F K' O
K
C B
Showing the multiplication K.K = F, the fourth power point
The circumcircle a^2yz + b^2zx + c^2xy = 0 X
X'
1
1. The product K.K = F, the square of the symmedian point is the fourth power point In the above figure the product K.K = F is illustrated. It is one of the easiest products, of course (other than using the centroid G, which acts as an identity). The circumconic in this case is the circumcircle with equation a2yz + b2zx + c2xy = 0 (1.1) The first conjugation sends K(a2, b2, c2) to K'(1/a2, 1/b2, 1/c2), the isotomic conjugate of K. The second is performed as follows, and is actually the isogonal conjugation. One draws AK' to meet the circumcircle at X', then draw the line parallel to BC through X' to meet the circumcircle again at X. Using BK', CK' similarly one obtains points Y, Z also on the circumcircle. Finally AX, BY, CZ are drawn and they concur at F. 2. The product I.I = K, the square of the incentre is the symmedian point
A
Z
Z' Y'
Y Mi I'
B
I K
The circumconic ayz + bzx +cxy =0 X'
X
Since the co-ordinates of I are I(a, b, c) the circumconic has equation 2
C
ayz + bzx + cxy = 0. The centre of this conic is the Mittenpunkt with co-ordinates Mi(a(b + c – a). b(c + a – b), c(a + b – c)).
(2.1)
(2.2)
The isotomic conjugation of I is I'(1/a, 1/b, 1/c) the lines AI', BI', CI' are now drawn to meet the circumconic in points X', Y, Z'. A line parallel to BC is drawn through X' to meet the circumconic at X. Points Y, Z are similarly obtained. Finally AX, BY, CZ are drawn and they concur at the symmedian point I.I = K. See the Figure on page 2. 3. Two more products In this Section we obtain two more products using the Spieker centre, for which the conic has equation (b + c)yz + (c + a)zx + (a + b)xy = 0. (3.1) The centre of this conic is the point X37, the crosspoint of the incentre and the centroid. See the figure on the next page. Using I we have I.Sp = X37. This is meant to be a joke showing that to obtain X37 by this method you have to know where it is already. You may check that the isotomic conjugate of I, the point I', has conjugate X37 = (a(b + c), b(c + a), c(a + b)) = I.Sp. When the symmedian point K is used instead of I the result is K.Sp = X42 the crosspoint of the incentre and symmedian point X42 (a2(b + c), b2(c + a), c2(a + b)). X42 lies on the line IG, where G is the centroid. This is also illustrated in the figure following.
3
A
Q' R
Y' Y
Z R'
Z'
Q X42
Line IG K'
I'
X37 G I
K (b + c)yz + (c + a)zx + (a + b)xy = 0
B
C
P'
P X'
X
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Article: CJB/2012/216
A Point, two Triangles and Two Conics Christopher Bradley Abstract: A triangle ABC is given along with its circumcircle Γ and any internal point P. Lines AP, BP, CP are extended to meet Γ at U, V, W respectively. Triangle UVW is drawn and their sides intersect internally in six points. BC meets AU at a, with b and c similarly defined. AU meets side VW at u with v and w similarly defined. It is shown in this article that a conic passes through the first six points and that a conic also passes through a, b, c, u, v, w.
A
W uc
ub
u
V
c b vc
wb
P O v
w
a B
va
wa
C
U Figure 1. Points U, V, W and conic ABCUVW Let P have co-ordinates (l, m, n). The line AP has equation ny = mz. This meets the circumcircle at the point U with co-ordinates (x, y, z), where x = – a2mn, y = m(b2n + c2m), z = n(b2n + c2m). (1.1) 1
The co-ordinates of V, W may be written down by cuclic change of x, y, z and a, b, c and l, m, n. For V we have x = l(c2l + a2n), y = – b2nl, z = n(c2l + a2n). (1.2) And for W we have x = l(a2m + b2l), y = m(a2m + b2l), z = – c2lm.
(1.3)
The conic ABCUVW is a circle and passes through the vertices of triangle of reference and so has the equation a2 yz + b2-zx + c2xy = 0. (1.4) 2. The twelve points created AU meets BC at a(0, m, n). b and c follow by cyclic change and have co-ordinates b(l, 0, n) and c(l, m, 0). The equation of UV is n(b2n + c2m)x + n(a2n + c2l)y – c2lmz = 0. Similarly the equations of VW and WU are l(c2l + a2n)y + l(b2l + a2m)z – a2mnx = 0.
(2.1)
(2.2)
and m(a2m + b2l)z + m(c2m + b2n)x – b2nly = 0.
(2.3)
UA^WV = u with co-ordinates x = l(2a2m + b2nl + c2lm), y = a2m2n, z = a2mn2. Points v, w have co-ordinates which may be obtained from those of u by cyclic change of x, y, z and a, b, c and l, m, n. UV^BC = wa has co-ordinates x = 0, y = c2lm, z = n(a2n + c2l) (2.4) Points ub, vc have co-ordinates which may be obtained from those of wa by cyclic change of x, y, z and a, b, c and l, m, n. WU^BC = va has co-ordinates x = 0, y = m(a2m + b2l), z = b2nl. (2.5) Points wb, uc have co-ordinates which may be obtained from those of va by cyclic change of x, y, z and a, b, c and l, m, n. 3. The conic va wa wb ub uc vc This has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0. 2
(3.1)
where we find u = 2a2m2n2(b2n + c2m) with v, w following from u by cyclic change of a, b, c, and l, m, n. And where we also find f = – lmn(a4mn + a2l(b2n + c2m) + 2b2c2l2, with g, h following from f by cyclic change of a, b, c, and l, m, n. This conic has centre P. Also if the six points involved are taken to be the vertices of entwining triangles (see the Figure) then the six internal points of intersection lie on a conic similar to this one. 4. The conic a b c u v w This conic also has the form of equation (3.1) where we find u = 2a2m3n3 with v.w following from u by cyclic change of a, b, c, and l, m, n. We also find f = – l3mn(b2n + c2m) with g, h following from f by cyclic change of a, b, c, and l, m, n. It was hoped to find the values of l, m, n for which these conics are circles, but the algebra involved was too difficult.
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Article: CJB/217/2012
An interesting Perspective in the Anticomplementary Triangle Christopher Bradley Abstract: A triangle ABC and its anticomplementary triangle A' B'C' are drawn and a point P is selected through which Cevians A'P, B'P, C'P are drawn meeting the sides of ABC in points L', M', N' and the sides of A', B', C' in points L, M, N. Circles LMN and L'M'N' are drawn meeting the sides of A'B'C' and ABC respectively in points U', V', W' and U, V, W. Several perspectives are created, but the most interesting is that of triangles A'B'C' and UVW with perspector Q.
L
U'
C'
A
W
To B'
R' V' M' N' S
R
V
P
B
L'
M
U W' N
A'
1
C
Q
1. The Anticomplementary Triangle A'B'C' and the Cevians A'P, B'P, C'P We take ABC to be the triangle of reference so that the line through A parallel to BC has equation y + z = 0. This is the line B'C'. Similarly the lines C'A' has equation z + x = 0 and A'B' has equation x + y = 0. It follows that A' has co-ordinates A'(– 1, 1, 1) and B', C' have co-ordinates B'(1, – 1, 1) and C'(1, 1, – 1). We now take P to have co-ordinates P(l, m, n). And it may now be shown that A'P has equation (m – n)x – (n + l)y + (l + m)z = 0. (1.1) Similarly B'P and C'P have equations (n – l)y – (l + m)z + (m + n)x = 0,
(1.2)
and (l – m)z – (m + n)x + (n + l)y = 0,
(1.3)
respectively. The line A'P meets BC at L' and B'C' at L. It may now be verified that L' has co-ordinates L'(0, l + m, n + l) and similarly M', N' have co-ordinates M'(l + m, 0, m + n) and N'(n + l, m + n, 0). The point L has co-ordinates L(2l + m + n, m – n, n – m) and similarly M, N have coordinates M(l – n, 2m + n + l, n – l) and N(l – m, m – l, 2n + l + m). 2. The circles L'M'N' and LMN Circles L'M 'N' and LMN have equations of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
(2.1)
We find that for L'M'N' u = (1/k)(m + n)(a2(l + m + 2n)(l + 2m + n) – (2l + m + n)(b2(l + m + 2n) + c2(l + 2m + n))), v = (1/k)(n + l)((2l + m + n)(b2(l + m + 2n) – c2(l + 2m + n)) – a2(l + m + 2n)(l + 2m + n)), w = – (1/k)(l + m)(a=(ll + m + 2n)(l + 2m + n) + (2l + m + n)(b2(l + m + 2n) – c2(l + 2m + n))), where (2.2) k = 2(l + m + 2n)(l + 2m + n)(2l + m + n). We also find that for LMN we have
2
u = (m + n)(m – n)(b2(l + m)(l + m + 2n) – c2(n + l)(l + 2m + n))/(l + m)(n + l)(l + m + 2n)(l + 2m + n), v = (l + n)(l – n)(a2(l + m)(l + m + 2n) – c2(m + n)(2l + m + n))/(l + m)(m + n)(l + m + 2n)(2l + m + n), w = (l + m)(l – m)(a2(l + n)(l + 2m + n) – b2(m + n)(2l + m + n))/(l + n)(m + n)(l + 2m + n)(2l + m + n). (2.3) Circle L'M'N meets BC at U with co-ordinates (x, y, z), where x = 0, y = a2(l + m + 2n)(l + 2m + n) + (2l + m + n)(b2(l + m + 2n) – c2(l + 2m + n)). z = a2(l + m + 2n)(l + 2m + n) – (2l + m + n)(b2(l + m + 2n) – c2(l + 2m + n)).
(2.4)
Circle L'M'N' meets CA at V with co-ordinates (x, y, z), where x = a2(l + m + 2n)(l + 2m + n) + b2(l + m + 2n)(2l + m + n) – c2(l + 2m + n)(2l + m + n), y = 0, z = – a2(l + 2m + n)(2l + m + n) + b2(l + m + 2n)(l + 2m + n) + c2(l + m + 2n)(2l + m + n). (2.5) Circle L'M'N' meets AB at W with co-ordinates (x, y, z), where x = – b2(l + m + 2n)(l + 2m + n) + c2(2l + m + n)(l + m + 2n) + a2(2l + m + n)(l + 2m + n), y = b2(2l + m + n)(l + m + 2n) + c2(2l + m + n)(l + 2m + n) – a2(l + m + 2n)(l + 2m + n), z = 0. (2.6) AL', BM', CN' are concurrent at a point S with co-ordinates S((l + m)(l + n), (m + l)(m + n), (n + m)(n + l)). (2.7) It follows since L'M'N' is a conic that AU, BV, CW are also concurrent at a point R. Circle LMN meets B'C' at a point U' with co-ordinates (x, y, z), where x = a2(l + m)(l + m + 2n)(n + l)(n + l + 2m), y = – (m + n)(2l + m + n)(b2(l + m)(l + m + 2n) – c2(l + n)(l + 2m + n)), z = (m + n)(2l + m + n)(b2(l + m)(l + m + 2n) – c2(l + n)(l + 2m + n)).
(2.8)
Points V' and W' have co-ordinates that may be obtained from those of U' by cyclic change of x, y. Z and a, b, c and l, m, n. A'U', B'V', C'W' meet at a point R', since A'L, B'M, C'N are concurrent and LMN is a conic. 3. The perspector Q of triangles UVW and A'B'C' That these two triangles should be in perspective was a surprise and indeed the main reason for performing the calculation and writing the article.
3
The co-ordinates of Q turn out to be (x, y. z), where x = (2l + m + n)(b2(l + m + 2n) + c2(l + 2m + n)) – 3a2(l2 + 3l(m + n) + 2m2 + 5mn + 2n2), y = a2(l + m + 2n)(l + 2m + n) – (2l + m + n)(3b2(l + m + 2n) – c2(l + 2m + n)) (3.1) 2 2 2 2 2 2 2 2 z = a (l + 3l(m + n) + 2m + 5mn + 2n ) + b (2l + l(3m + 5n) + m + 3mn + 2n ) – 3c2(2l2 + l(5m + 3n) + 2m2 + 3mn + n2). One cannot be other than mystified at the lack of symmetry of the co-ordinates.
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Article: CJB/218/2012
Properties of the Triangle of Excentres Christopher Bradley X
To X
I3
A
I2 O+ O K+I
K H
To Y
C
B
To Z
Z I1
1
Abstract: In the triangle of excentres the orthocentre is the incentre I of ABC, the Symmedian point K+ is the Mittenpunkt of ABCand the nine-point centre is the circumcentre O of ABC. The circumcentre O+ lies on the line OI and is such that O+O = OI. These properties are proved as also the fact that KI passes through K+. 1. Circle I1I2I3 and its tangents The co-ordinates of the excentres are I1(– a, b, c), I2(a, – b, c), I3(a, b, – c). It is easy to check that the equation of the circle I1I2I3 is a2yz + b2zx + c2xy +(x + y + z)(bcx + cay + abz) = 0. (1.1) The tangents to this circle at the points I1, I2, I3 have equations I1: (b + c)x + a(y + z) = 0, I2: (c + a)y + b(z + x) = 0, I3: (a + b)z + c(x + y) = 0.
(1.2) (1.3) (1.4)
Tangents at I2, I3 meet at X with co-ordinates X(– a(a + b + c), b(a + b – c), c(a – b + c)). Tangents at I3, I1 meet at Y with co-ordinates Y(a(a + b – c), – b(a + b + c),c(b + c – a)). Tangents at I1, I2 meet at Z with co-ordinates Z(a(c + a – b), b(b + c – a), – c(a + b + c)). 2. The symmedian point K+ The symmedian point K+ of triangle I1I2I3 is the point of concurrence of lines XI1, YI2, ZI3 whose equations are respectively (c – b)x = a(y – z), (a – c)y = b(z – x), (b – a)z = c(x – y). (2.1) The co-ordinates of K+ are therefore (a(b + c – a), b(c + a – b), c(a + b – c)). The co-ordinates will be recognized as those of the Mittenpunkt Mi, X9 in Kimberling’s list of Triangle Centres. It is a well known fact that in triangle I1I2I3 that II1 is perpendicular to I2I3 etc. Hence the incentre I is the orthocentre H+ of triangle I1I2I3. From the usual formulae for the centre of a conic we find the co-ordinates of O+ to be (x, y, z), where x = a(a3 + a2(b + c) – a(b + c)2 – b3 + b2c + bc2 – c3) (2.2)
2
with y, z following by cyclic change of a, b, c. It may now be checked that this is the point on IO such that O+O = OI and hence the nine-point centre of triangle I1I2I3 is the circumcentre O of triangle ABC. The equation of the line IK is bc(c – b)x + ca(a – c)y + ab(b – a)z = 0 and it may be checked that K+ lies on this line. Finally, after some heavy algebra, it may be checked that K+ lies on the conic I1I2I3OH.
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(2.3)
Article: CJB/219/2012
The Twelve point Circle Christopher Bradley Abstract: A construction is given showing how to convert the nine-point circle to a twelve point circle, with its twelve points lying on the sides of two triangles.
A
D' C'
U
F N O
E
H
T
E'
W
V F' B
B'
M
L
D
A'
1
C
Description of the Construction In triangle ABC the orthocentre is H, the circumcentre is O and U, V, W are the midpoints of AH, BH, CH respectively. Lines are drawn through U, V, W parallel to BC, CA, AB respectively. These lines define a triangle A'B'C'. The nine-point circle, centre T the midpoint of OH, is defined by points U, V, W and it passes through the midpoints L, M, N of the sides and points D, E, F the feet of the altitudes AH, BH, CH respectively. The symmetry of the figure shows that the midpoint of each of the lines AA', BB', CC' is T the centre of the nine-point circle. It follows that circle UVW is also the nine-point circle of triangle A'B'C'. The midpoints of the sides of A'B'C' are the points U, V, W. The orthocentre of A' B'C' is the point O and the feet of the perpendiculars A'O, B'O, C'O are the points D', E', F' and L, M, N are the midpoints of the segments A'O, B'O, C'O. The common nine-point circle now becomes the twelve-point circle, the twelve points being U,V,W, D, E, F, L, M, N, D', E', F'. It is clear that a conic passes through A, B, C, A', B', C' centre T.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB /220/2012
A Conical Hexagon with Main Diagonals Concurrent Christopher Bradley Abstract: A hexagon ABCDEF is inscribed in a conic with diagonals AD, BE, CF concurrent at a point P. It is shown that this property is replicated and that both conics have the same polar with respect to P. This polar is one of the Pascal lines.
1
5 15 46
F
3 E
A
35 4
P D 26 \B
C
13
2
24
To 6
1. Introduction A hexagon ABCDEF is inscribed in a conic Σ with diagonals AD, BE, CF concurrent at a point P. Projective co-ordinates are used. It is shown that AB^DE = 1, BC^EF = 2, CD^FA = 3 lie on a line, which is the polar of P with respect to Σ. Point BC^AF, AB^CD, BC^DE, CD^EF, AF^DE, EF^AB are shown to lie on another conic Γ that also has the property that three diagonals pass through P and that the polar of P with respect to Γ is also the line 123. The line is, of course, a Pascal line of both hexagons, but that is incidental to the other properties. An infinite sequence of such conics does, of course, exist.
1
2. The co-ordinates of points We take Σ to have equation y2 = zx and without generality we may take points A → F to have co-ordinates: A(1, 0, 0), B(1, 1, 1), C0, 0, 1), D(d2, d, 1), E(e2, e, 1), F(f2, f, 1). (2.1) We now find the value of E so that AD, BE, CF are concurrent at a point P. The equation of AD is y = dz and the equation of CF is x = fy. These met at a point P with coordinates P(fd, d, 1). The equation of BE is x – (1 + e)y + ez = 0. (2.2) The condition that P lies on BE is e = d(f – 1)/(d – 1).
(2.3)
Hence E has co-ordinates (x, y, z), where x = d2(f – 1)2, y = d(f – 1)(d – 1), z = (d – 1)2.
(2.4)
3. The equations of the sides of ABCDEF These are AB; BC: CD: DE: EF: FA:
y = z, y = x, x = dy, (d – 1)x – d(d + f – 2)y – d2(1 – f)z = 0, (d – 1)x + (f – 2df + d)y + df(f – 1)z = 0, y = fz.
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
4. The polar line of P The points of this line are 1, 2, 3 where 1 = AB^DE: 2 = BC^EF: 3 = CD^FA:
x = d(2 – f), y = 1, z = 1, x = fd, y = fd, z = 2d – 1, x = fd, y = f, z = 1.
(4.1) (4.2) (4.3)
x – 2dy +dfz = 0
(4.4)
The equation of 123 is
It is easily checked that 123 is the polar of P with respect to Σ.
2
5. The second conic Γ We now show that six points of intersection of the sides of ABCDEF lie on a conic Γ. These points have co-ordinates as follows: AB^CD = 13: x = d, y = 1, z = 1, (5.1) 2 AB^EF = 15: x = d(f – 3f + 1) + f, y = 1 – d, z = 1 – d, (5.2) BC^FA = 26: x = f, y = f, z = 1, (5.3) 2 DE^FA = 46: x = d(d + f – 2f), y = f(d – 1), z = d – 1, (5.4) 2 2 2 BC^DE = 24: x = d (f – 1), y = d (f – 1), z = d + df – 3d + 1, (5.5) 2 2 CD^EF = 35: x = d f(1 – f), y = df(1 – f)z = d – 2df + f. (5.6) These points lie on a conic Γ of the form ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy, where, u = 2(1 – d), v = 2d(d – f), w = 2d2f(f – 1), l = 2df(1 – d), m = d2 + df – df2 – f, n = 2d(f – 1).
(5.7)
(5.8)
It may now be shown that the lines 15 24, 26 35 and 13 46 all pass through P and that line 123 is also the polar of P with respect to Γ.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
3
Article: CJB/2012/221
Four Special Conical Hexagons all with same Polar Christopher Bradley Abstract: Given a hexagon ABCDEF inscribed in a conic with AD, BE, CF concurrent at P, the tangents tA, tB, tC, tD, tE, tF are drawn. Points jk = tj^tk are determined and it is shown that 12, 23, 34, 45, 56, 61 lie on a conic. Lines 61 12, 34 45 meet at point 1, lines 12 23, 45 56 meet at 2 and lines 23 34, 56 61 meet at 3. Points 1, 2, 3 are collinear and it is shown that 123 is the polar of P with respect to both conics.
51 62
M
61 G A
12 13
F
B P
23
L 56
E H
C
46
To 1
D
45 K
34
24 J
35
3
2
The Four Conical Hexagons Abstract continued: Points 4 = AB^DE, 5 = BC^EF, 6 = CD FA all lie on line 123. Further points are defined as follows: BC^FA = G, CD^AB = H, DE^BC = J, EF^CD = K, FA^DE = L, AB^EF = M. It is then shown that GHJKLM is a conic and GK, HL, JM all pass through P, the polar of P with respect to this ellipse also being the line123. Finally six more points may be determined from the tangents, which again have the same properties involving P and its polar 123. 1
1. The conic ABCDEF and the point P We use homogeneous projective co-ordinates and take the conic ABCDEF to have equation y2 = zx. (1.1) We take points on this conic to have co-ordinates as follows: A(1, 0, 0), B(`1, 1, 1), C(0, 0, 1), D(d2, d, 1), E(e2, e, 1), F(f2, f, 1). The equations of AD, BE, CF are AB: BE: CF:
y = dz, x – (1 + e)y + ez = 0, x = fy.
(1.2)
(1.3)
These three lines meet at P(fd, d, 1) provided e = d(f – 1)/(d – 1),
(1.4)
and then E has co-ordinates E(d2(f – 1)2, d(f – 1)(d – 1), (d – 1)2)…
(1.5)
2. The tangents at A, B, C, D, E, F These have equations as follows: tA: tB : tC : tD: t E; tF:
z = 0, 2y = x + z, x = 0, 2dy = x + d2z, 2d(d – 1)(f – 1)y = (d – 1)2x + d2(f – 1)2z, 2fy = x + f2z.
These meet in consecutive pairs at points whose co-ordinates are (x, y, z), where 12 = tA^tB: x = 2, y = 1, z = 0, 23 = tB^tC: x = 0, y = 1, z = 2, 34 = tC^tD: x = 0, y = d, z = 2, 45 = tD^tE: x = 2d2(f – 1), y = d(d + f – 2), z = 2(d – 1), 56 = tE^tF: x = 2df(f – 1), y = d(2f – 1) – f, z = 2((d – 1), 61 = tF^tA: x = 2f, y = 1, z = 0. It may now be checked that point 12, 23, 34, 45, 56, 61 lie on a conic with equation ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, where u = 2d, v = 8df, w = 2d2f, l = – 2df (d + 1), m = 2d2 + df(f – 2) + f, n = – 2d(f + 1) . 2
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
(2.7) (2.8) (2.9) (2.10) (2.11) (2.12)
(2.13) (2.14)
3. Points 1, 2, 3 and the polar line and points 4, 5, 6 The line 23 34 has equation x = 0, and the line 56 61 has equation x = f(2y – fz). These meet at the point 3 with co-ordinates (0, f, 2). The line 12 23 has equation x – 2y + z = 0 and the line 45 56 has equation (d – 1)2x + d(1 – f)(2(d – 1)y + d(1 – f)z) = 0. These meet at the point 2 with co-ordinates (2d(f – 1), df – 1, 2(d – 1)).
(3.1) (3.2)
The line 34 45 has equation x = d(2y – dz) and line 61 12 has equation z = 0. These meet at the point 1 with co-ordinates (2d, 1, 0). It may now be checked that point 1, 2, 3 are collinear on the line x = d(2y – fz).
(3.3)
And finally that 123 is the polar of the point P(fd, d, 1) with respect to both conics.
(3.4)
An interesting feature is that 12 45, 23 56 and 34 61 all pass through P. The line AB has equation y = z and the line DE has equation (d – 1)x – d((d + f – 2)y + d(1 – f)z). These meet at point 4 with co-ordinates (d(2 – f), 1, 1).
(3.5)
The line BC has equation x = y and the line EF has equation (d – 1)x + (f – d(2f – 1))y + df(f – 1)z = 0. These two lines meet at point 5 with co-ordinates (fd, fd, 2d – 1).
(3.6)
The line CD has equation x = dy and FA has equation y = fz. These two lines meet at point 6 with co-ordinates (df, f, 1). It may now be shown that points 4, 5, 6 all lie on line 123. 4. Points G, H, J, K, L, M and the conic on which they lie Lines BC and FA meet at G(f, f, 1). Lines CD and AB meet at H(d, 1, 1). Lines DE and BC meet at J(d(d + f(f – 2)), f(d – 1), d – 1). Lines EF and CD meet at K(d2f(1 – f), df(1 – f), d2 – 2df + f). Lines FA and DE meet at L(d(d + f(f – 2)), f(d – 1), d – 1). AB and EF meet at M(d(f2 – 3f + 1) + f, 1 – d, 1 – d). These six points lie on the conic with equation of the form (2.13), with 3
u= 2(1 – d), v = 2d(d – f), w = 2d2f(f – 1), l = 2df(1 – d), m = d2 + df(1 – f) – f, n = 2d(f – 1). Finally it may be shown that the polar of P with respect to this conic is again the line 123. 5. Six more points and another conic The tangents tA etc. are listed at the beginning of Section 2. Six pairs of intersection lead to six more points. These are tF^tB = 62 (2f, f + 1, 2), tE^tA = 51 (2d(f – 1), d – 1, 0), tA^tC = 13 (0, 0, 1), tB^tD = 24 (2d, d + 1, 2), tC^tE = 35 (0, d(f – 1), 2(d – 1)), tD^tF = 46(2df, d + f, 2). Theses six points lie on a conic with equation of the form (2.13) with u = 2(d – 1), v = 0, w = 2d2f(1 – f), l = 2d(d – 1)f, m = (f + df2 – 2d2), n = 2d(1 – f). And once again the polar of P with respect to this conic is the line 123.
Flat4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article: CJB/222/2012
The Missing Point on the Euler Line Christopher Bradley
A
OS = 0.6109790663 cm OH = 2.4439162652 cm
F V
N
M
W
S
U
B
E
H
O T
L
D
C
Result: 0.2500000000= OS/OH S is the missing point
Abstract: Given a triangle ABC, circumcentre O, orthocentre H, let the midpoints of sides be L, M, N and suppose AO, BO, CO meet the sides BC, CA, AB respectively at U, V, W then the centre of the ellipse LMNUVW is a point S on the Euler line such that OH = 4OS. There appears to be no reference to this point in the literature, and so we call it ‘The Missing Point’. 1. The theory of the ellipse LMNUVW We use areal co-ordinates throughout with side lengths a, b, c. The midpoints L, M, N obviously have co-ordinates L(0, ½, ½), M(½ ,0, ½), N( ½, ½, 0).
1
The circumcentre has un-normalized co-ordinates O(a2(b2 + c2 –a2), b2(c2 + a2 – b2), c2(a2 + b2 – c2)).
(1.1)
To obtain normalized co-ordinates one must divide by 2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4.
(1.2)
The equation of AO is c2(a2 + b2 – c2)y = b2(c2 + a2 – b2).
`
This meets BC, x = 0, at U with co-ordinates (x, y, z), where x = 0, y = b2(c2 + a2 – b2), z = c2(a2 + b2 – c2).
(1.3)
(1.4)
Points V and W have co-ordinates which may be obtained from those of U by cyclic change of x, y, z and a, b, c. Now that we have the co-ordinates of the six points, we may now find the equation of the conic passing through them (which must be an ellipse). It s equation is of the form ux2 |+ vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (1.5) where we find u = 2b2c2(a2 + b2 – c2)(c2 + a2 – b2), v = 2c2a2(b2 + c2 – a2)(a2 + b2 – c2), w = 2a2b2(c2 + a2 – b2)(b2 + c2 – a2), f = – a2(b2 + c2 – a2)(a2(b2 + c2) – (b2 – c2)2), g = – b2(c2 + a2 – b2)(b2(c2 + a2) – (c2 – a2)2), h = – c2(a2 + b2 – c2)(c2(a2 + b2) – (a2 – b2)2).
(1.6) (1.7) (1.8) (1.9) (1.10) (1.11)
2. The centre S of this ellipse and its location The centre of a conic with equation of the form (1.5) has centre at (x, y, z), where x = vw – gv – hw – f2 + fg + hf, y = wu – hw – fu – g2 + gh + fg, z = uv – fu – gv – h2 + hf + gh.
(2.1) (2.2) (2.3)
Using values from Section 1 we find that S has un-normalized co-ordinates (x, y, z), where x = 2a4 + (b2 – c2)2 – 3a2(b2 + c2), (2.4) 4 2 2 2 2 2 2 y = 2b + (c – a ) – 3b (c + a ), (2.5) 4 2 2 2 2 2 2 z = 2c + (a – b ) – 3c (a + b ). (2.6) It may now be shown that this point is located at S on the Euler line, where OH = 4OS. 2
As far as we can determine, this point which lies half way between the circumcentre and the nine-point centre, has not so far been listed with any properties, and as such we have termed it the missing point!
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
Article: CJB/223/2012
When a non-regular Cyclic Pentagon leads to Another Christopher Bradley Abstract: It can be arranged that a non-regular cyclic pentagon ABCDE, by special adjustment of D and E, produces a second cyclic pentagon A'B 'C'D'E', where A' = BD^CE, B' = AC^DE, C'= AB^DE, D'= AB^CE and E' = AC^BD.
E'
D' A B
C O O'
C'
D
E
A'
Adjusted Figure 1. The cyclic non-regular pentagon ABCDE
1
B'
We take an obviously convenient arrangement of points A, B, C, D, E with co-ordinates A(0, 1), B( – ½, √3/2), C( ½, √3/2), D((e2 – 1)/(e2 + 1), – 2e/(e2 + 1)), E((1 – e2)/(1 + e2), – 2e/(1 + e2)), where the value of e must be positive and will eventually be determined to ensure A'B'C'D'E' is also cyclic with an equation of the form x2 + y2 + 2fy + k = 0, (1.1) where f and k will be determined in terms of e. It is immediately possible to determine the equations of the sides of ABCDE. These are: AB: (1 – √3)x + (1 + √3)y = 1 + √3, (1.2) AC: (√3 – 1)x + (√3 + 1)y = – ( 1 + √3), (1.3) BD: (e + √3)x + (e√3 – 1)y = e – √3 (1.4) CE: (e + √3)x + (1 – e√3)y = √3 – e, (1.5) 2 DE: (1 + e )y = – 2e. (1.6) 2. The pentagon A'B'C'D'E' The co-ordinates of A', B', C', D', E' may now be obtained and are: A' = BD^CE: x = 0, y = (e – 3)/(e√3 – 1), B' = AC^DE: x = (√3 + 2)(e + 1)2/(1 + e2), y = – 2e/(1 + e2), C' = AB^DE: x = – (√3 + 2)(e + 1)2/(1 + e2), y = – 2e/(1 + e2), D' = AB^CE: x = (e + 1)/(e(√3 – 1) + √3 + 1)), y = (e + 3)/(e(√3 – 1) + √3 + 1)), E' = AC^BD: x = – (e + 1)/(e(√3 – 1) + √3 + 1)), y = (e + 3)/(e(√3 – 1) + √3 + 1)),
(2.1) (2.2) (2.3) (2.4) (2.5)
3. When the pentagon A'B'C'D'E' is cyclic Obviously if A'B'C'D'E' is cyclic its origin must be on the y-axis and so it must have an equation of the form x2 + y2 + 2fy + k = 0. (3.1) Putting two of the co-ordinates of points into this equation we may find f and k in terms of e and a third co-ordinate then allows e to be determined numerically. (Two of the points do not prove anything new.) The values of f and k do not need to be written down, but we give the numerical values of the constants to ten significant places (exact values are possible in terms of trigonometrical expressions). 2
The values are e = 0.6031557384, f = 12.05597092, k = – 28.91984973.
Flat 4, Terrill Courty, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
(3.2) (3.3) (3.4)
Article: CJB/224/2012
The Sixty Pascal Poles Christopher Bradley Abstract: When six points lie on a conic then sixty Pascal lines may be constructed. In this paper it is shown how the poles of sixty these lines may be constructed without drawing tangents. Their properties are, of course, the duals of those of the Pascal lines.
To BF^CE
E'
F A O P
B
E
D C D'
C'
A Pascal Line and its Pole P 1
1. Introduction In the figure points A, B, C, D, E, F are drawn in that order on a conic (drawn as a circle, which is sufficiently general). Points C'= AB^DE, D' = BC^EF, E' = CD^AF (that is, intersections of pairs of opposite sides are drawn) and points C'D'E' is a straight line, being the Pascal line of the ordered six points. Other orders produce altogether 60 Pascal lines. These lines have poles with respect to the defining conic and we show that the pole P of the Pascal line defined above lies on each of the three lines AD^CF AC^DF, BF^CE BE^CF, BD^AE AD^BE. The 60 poles satisfy dual properties to those of the 60 lines, and as their properties are seldom listed we mention some of these properties in the concluding section. 2. The Pascal line for the hexagon ABCDEF This is the line C'D'E' where C' = AB^DE, D' = BC^EF, E' = CD^AF. We take the circle to have equation x2 + y2 = 1 with point A having co-ordinates ((1 – a2)/(1 + a2), 2a/(1 + a2)) and similarly for B, C, D, E, F with parameters b, c, d, e, f. The equation of AB is well-known to be (1 – ab)x + (a + b)y = (1 + ab).
(2.1)
with similar equations for other lines. The co-ordinates of C' = AB^DE may now be calculated and are (x, y), where x = – (abd + abe – ade – bde – a – b + d + e)/(abd + abe – ade – bde + a + b – d – e), (2.2) y = 2(ab – de)/(abd + abe – ade – bde + a + b – d – e). (2.3) The co-ordinates of D' and E' follow by appropriate changes of parameters. The equation of the Pascal line is thus mx + ny = 1, (2.4) where m = – (a(b(c(d – f) + ef – 1) – f(de – 1)) + bc(1 – de) + cd(ef – 1) + e(d – f)) all divided by (a(b(c(d – f) + ef + 1) – f(de + 1)) – bc(1 + de) + cd(ef + 1) – e(d – f)). n = a(b(d + e) + c(d – f) – d(e + f)) – b(c(e + f) + e(d – f)) + cf(d + e) all divided by (a(b(c(d – f) + ef + 1) – f(de + 1)) – bc(1 + de) + cd(ef + 1) – e(d – f)). 3. The pole P and three lines through P
2
(2.5)
(2.6)
The pole P, of course, has co-ordinates (m, n), but it cannot be easily located without introducing lines through P that may be constructed. This can be done by drawing appropriate tangents to the circle from points on the Pascal line, but that is not satisfactory and indeed one can do better. We prove in fact that the lines (i) BF^CE BE^CF, (ii) BD^AE BE^ AD and (iii) AD^CF AC^DF all pass through P, so that P may be located without introducing any additional points. In fact instead of producing 60 Pascal lines and their properties it would have been possible to produce 60 poles with consequential properties that are the duals of those properties. It is simply that Pascal found the line and not the point. We now deal with (i). The equations of BE and CF are of the form (2.1) with parameter (b, e) and (c, f) rather than (a, b). Their point of intersection BE^CF has co-ordinates that may be obtained from (2.2) and (2.3) with e, c, f replacing a, d, e. The co-ordinates of BF^CE may now be obtained from those of BE^CF by interchange of e and f. The equation of the line joining these two points is now px + qy = 1, (3.1) where p = – (b(c(e + f) – ef – 1) – c(ef + 1) + e + f)/(b(c(e + f) – ef + 1) + c(1 – ef) – e – f), and q = 2(bc – ef)/(b(c(e + f) – ef + 1) + c(1 – ef) – e – f).
(3.2) (3.3)
It may now be checked that P(m, n) lies on this line. By symmetry it now follows that P also lies on lines (ii) and (iii). Thus P may be located by using only the initial points A, B, C, D, E, F. And to each Pascal line there is a corresponding pole. 4. Some properties of the 6o poles Corresponding to the 20 Steiner points (through any one of which 3 Pascal lines pass) we have (i) 20 (Steiner) lines each of which has 3 poles lying on it; Corresponding to the 60 Kirkman points (that have the property that they lie 3 at a time on lines through the 20 Steiner points) we have (ii)
60 (Kirkman) lines such that there are 20 points of concurrences with 3 Kirkman lines through each such point and these 20 points lie one each on the 20 Steiner lines.
Similar dual correspondences exist involving analogues of the 20 Cayley lines, the 15 Plucker lines and the 15 Salmon points.
3
And all this occurs without mentioning Brianchon’s theorem involving six tangents to a conic rather than six points on a conic as in Pascal’s theorem.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
Article : CJB/225/2012
A Typical Pascal line Drawing Christopher Bradley
Y To B'
A B
Y1 Z2
Y' P X' X1
C
D
C'
X
A'
1
X2 Z' Y2
F Z1
To Z E
Description ABCDEF is a cyclic hexagon (though the results remain true if the circle is replaced by a conic). A'B'C' is a Pascal line, where A' = BC^AF, B' = CD^AF and C' = DE^AB. We denote this by (AEC, DBF), where care must be taken with the order of letters. Points X, Y, Z are given by X = AC^DF, Y = AE^DB, Z = EC^BF. Points X', Y', Z' are given by X' = AD^CF, Y' = AD^EB, Z' = CF^EB. In Article 225 it is shown that XX', YY', ZZ' are concurrent at a point P which is the pole of A'B'C' with respect to the circle. Points X1, X2 are given by X1 =CE^BD, X2 = AE^BF and X'X1X2 is the Pascal line (ABC, FED). Points Y1, Y2 are given by Y1 = AC^FB, Y2 = EC^DF and Y'Y1Y2 is the Pascal line (AEF, BDC). Points Z1, Z2 are given by Z1 + AE^FD, Z2 + CA^DB and Z'Z1Z2 is the Pascal line (BFA, CED) The poles of the three additional Pascal lines can easily be constructed, but then the diagram would be too crowded to be easily observed.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/226/2012
Special Conical Hexagons Christopher Bradley Abstract: A Hexagon ABCDEF inscribed in a Conic is said to be ‘special’ if AD, BE, CF are concurrent at a point P. When this happens hexagons formed by taking the intersections of neighbouring tangents are proved also to be special and conical with the same point P.
fb ea
fa F
A
ef
E ab
B ac
df
P
C
D
de
cd
bc
ce
bd
be
ad 1. Introduction
We take the inner conical hexagon ABCDEF (without loss of generality) to lie on a circle. The tangents at A and B meet at ab and in general the tangents at X and Y ≠ X meet at xy. We prove that ab de, bc ef and cd fa all pass through P and Cabri shows the six points lie on a conic and thereby form a special conical hexagon. It is also shown that ca bd, fb ce and ea bd all pass through P and that Cabri again shows these six points also form a special conical hexagon. 2. The points We take points on the unit circle to have parameters a, b, c, d, e, f and co-ordinates of the form x = (1 – a2)/(1 + a2), y = 2a/(1 a2). (2.1) 1
cf
The tangent at A has equation (1 – a2x + 2ay = (1 + a2),
(2.2)
and similarly for other tangents with appropriate change of parameter. Since AD, BE and FC meet at a point P then P has co-ordinates x = – (a(b(d – e) + de – 1) + b(1 – de) – d + e)/ (a(b(d – e) + de + 1) – b(1 + de) + d – e), y = 2(ad – be)/(a(b(d – e) + de + 1) – b(1 + de) + d – e), (2.3) with the condition for concurrence being abd – abe – acd + acf + ade – adf + bce – bcf – bde + bef + cdf – cef = 0.
(2.4)
The tangents at A and B meet at the point ab with co-ordinates x = (1 – ab)/(1 + ab), y = (a + b)/(1 + ab),
(2.5)
and similarly for other points de, etc. The equation of the line ab de is (a(b(d + e) – de – 1) – b(de + 1) + d + e)x + 2(de – ab)y +a(b(d + e) – de + 1) + b(1 – de) – d – e = 0. (2.6) Other lines in the figure have similar equations with appropriate changes of parameters. It may be checked from the equations of lines such as (2.6) by virtue of Equations (2.3) and (2.4), that they all pass through P. Cabri indicates that ab fa ef de cd bc forms a conic as do the points fb ea df ce bd ac. The equations of these conics are technically very difficult and their form extremely lengthy and it is too difficult to give these equations without printing error.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/227/2012
The Remarkable Eight-Point Conic Christopher Bradley
Q
To Q H V A
G
B
S To R
P
To U
D
R F
C
E T
U
Abstract: Two conics intersect at A, B, C, D and AC meets BD at P. A line through P is drawn meeting one of the conics at F, G and the other at E, H. Tangents are drawn at F, G, E, H. Tangents at F and G meet at Q and tangents at E and H meet at R. Tangents at H and G meet at V, tangents at H and F meet at S, tangents at E and F meet at T and tangents at E and G meet at U. A conic now passes through the 8 points A, B, C, D, S, T, U, V and the line QR is the polar of P with respect to all three conics. The general case is technically too difficult to establish, so here we provide a numerical case. 1. The Starting Configuration We use areal co-ordinates throughout. We take the two conics to have equations yz + zx + xy = 0, 1
(1.1)
and 12yz + 10zx + 5xy = 0.
(1.2)
Their four common points are A(1, 0, 0), B(0, 1, 0), C(0, 0, 1), D(14, – 10, 35). The intersection of AC and BD is the point P(2, 0, 5). A line is now drawn through P with equation 5x + 6y – 2z = 0.
(1.3)
This meets conic (1.1) at points H(– 6, 10, 15), E(2, – 1, 2), and it meets conic (1.2) at points F(– 4, 5, 5) and G(18, – 11, 15). 2. The 8-point conic and the polar line The tangent at F has equation 15x + 8y + 4z = 0, the tangent at G has equation 10x + 27y + 6z = 0. The tangent at E has equation x + 4y + z = 0 and the tangent at H has equation 25x + 9y + 4z = 0. The tangents at F and G meet at R(– 1, – 3, 13) and the tangents at E and H meet at Q(– 12, – 10, 65). The tangents at H and F meet at S(4, – 40, 65), the tangents at H and G meet at V(54, 110, – 585). The tangents at E and F meet at T(8, 11, – 52), the tangents at E and G meet at U(3, – 4, 13). It may now be checked that A, B, C, D, S, T, U, V all lie on the conic with equation 6yz – 44zx – 169xy = 0.
(2.1)
The equation of QR is 5x + 7y + 2z = 0.
(2.2)
It may be shown easily that this is the polar of P(2, 0, 5) with respect to each of the three conics.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP
2
Article: CJB/228/2012
How 2 Conics through 4 Points generate 6 more such Conics Christopher Bradley
To 67 68 Conic 57 68 5
G
3 57
8 6
1
58 58 67 Conic
Conic 13 24
7 13
14
D
A
E To 12
F
34 4
78 C
56
24 2 23 Conic 14 23
B
Conic 56 78
Conic 12 34
Description Two Conics through 4 point A, B, C, D are drawn (in the figure the light blue and the orange). Their tangents to the first (the orange) at A, B, C, D are labelled 1, 2, 3, 4 and the tangents to the second (the light blue) are labelled 5, 6, 7, 8. Intersections of these tangents are labelled 12, 13, 14, 23, 24, 34 and 56, 57, 58, 67, 68, 78. It is now found that six 6 point conics may be drawn: (i) ABCD 13 24; (ii) ABCD 12 34; (iii) ABCD 14 23; (iv) ABCD 56 78; 1
(v) (vi)
ABCD 57 68; ABCD 58 67.
Care must be taken over the choice of D. For example, if 24 lies on AC then Conic (i) degenerates into two straight lines AC BD and similarly for other Conics. It is also found that 12 34 56 78 lie on the line joining the diagonal points EF = AC^BD AD^BC of the quadrilateral ABCD with similar sets of points on FG and GE. With six more conics through A, B, C, D the method may be continued indefinitely providing a countable number of conics through the four starting points.
Flat 4 Terrill Court 12-14 Apsley Road BRISTOL BS8 2SP
2
Article: CJB/229/2012
When a Conical Quadrilateral produces two 6-point Conics Christopher Bradley Abstract: If a quadrilateral ABCD is drawn in a Conic and the tangents are drawn at A, B. C. D then the sides of ABCD and the intersections of the tangents produce two six point conics and a polar line.
U
R
A
D
S E
Q
G
C B
P T
F
1. Introduction Let ABCD be a quadrilateral inscribed in a conic and suppose that tangents are drawn at A, B, C, D to the conic. Tangents at A and B meet at S. Tangents at B and C meet at P. Tangents at C and D meet at Q. Tangents at D and A meet at R. In this short article we show that a conic passes through A, B, C, D, P and R and that another conic passes through A, B, C, D, Q and S. The configuration and proof also apply to Article 228. 2. The starting conic and the tangents We use areal co-ordinates throughout with ABC as triangle of reference and D as the point (2, – 1, 2). The most general conic through these points has equation 1
(1 + k)yz + zx + (1 – k)xy = 0,
(2.1)
where k is a constant not equal to 1, – 1. These two exceptions prevent conic (2.1) being degenerate. The equations of BC, CA, AB are respectively x = 0, y = 0, z = 0. The equation of AD is 2y + z = 0, that of BD is x = z and that of CD is x + 2y = 0. Point E = BD^AC has co-ordinates E(1, 0, 1). Point F = AB^CD has co-ordinates F(– 2, 0, 1) and the point G = AD^BC has co-ordinates G(0, 1, – 2). The equation of the tangent at (f, g, h) to the conic with equation (2.1) is (1 + k)(hy + gz) + hx + fz + (1 – k)(gx + fy) = 0.
(2.2)
Thus the tangents at A, B, C, D have equations: tA: y(1 – k) + z = 0, tB : x(1 – k) + z(k + 1) = 0, tC : x + y(k + 1) = 0, tD: x(k + 1) + 4y + z(1 – k) = 0.
(2.3) (2.4) (2.5) (2.6)
3. Resulting points Co-ordinates of the following points now are: S = tA^tB ; (k + 1, 1, k – 1), P = tB^tC : (– (1 + k), 1, 1 – k), Q = tC^tD : (k + 1, – 1, k + 3), R = tD^tA : (k – 3, 1, k – 1), U = tA^tC : (k + 1, – 1, 1 – k), T = tB^tD : ( – (1 + k), k, 1 – k).
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
4. The Two Conics
It may now be checked that a conic with equation (3 – k)(k + 1)yz + 2zx + (k – 1)2 xy = 0
(4.1)
passes through A, B.C, D, P, R and that a conic with equation (k + 1)2yz + 2zx + (1 – k)(k + 3)xy = 0
(4.2)
passes through A, B, C, D, Q, S.
2
5. Three lines It may also be checked that S, E, Q, G are collinear with equation – x + 2y + z = 0,
(5.1)
and that F, P, E, R are collinear with equation x + 2y – z = 0,
(5.2)
and that U, G, T, F lie on the polar of E with respect to all three conics with equation x + 2y + z = 0.
(5.3)
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3
Article: CJB/230/2012 A Quadrilateral and the Conics and Points arising Christopher Bradley
U
13
34
41 S A X P 24
E B
D
42
R
Y 23
C
F
Q 31 12 T
G
Abstract: ABCD is an arbitrary quadrilateral inscribed in a general conic. The tangents at its vertices form a second quadrilateral PQRS. There emerges the polar line FGTU and 8 points of intersection of ABCD and the tangential quadrilateral PQRS. Many conics result, but two of them in particular meet on the diagonal PR at points X, Y in the figure. X and Y appear to have properties as important as any of the 8 vertices of the quadrilaterals and the four points on the polar line. Not by any means are all the conics that can be drawn featured in this article and a second article is envisaged. What determines why the diagonals PR and QS have such different properties are (as far as we are concerned) open and challenging questions. 1. Defining the initial points and lines 1
We use areal co-ordinates throughout with ABC as triangle of reference. We take D to have the numerical co-ordinates D(2, – 1, 2) and the conic ABCD to have the most general equation containing A, B, C and D, which is (1 + k)yz + zx + (1 – k)xy = 0, (1.1) where k is an arbitrary constant not equal to 1 or – 1. The sides and diagonals of ABCD have the following equations; AB: z = 0; BC: x = 0; CA: y = 0; AD: 2y + z = 0; BD: z = x; CD: 2y + x = 0.
(1.2)
The diagonal points have co-ordinates: E = BD^AC: (1, 0, 1); F = AD^BC: (0, 1, – 2); G = AB^CD: (– 2, 1, 0).
(1.3)
The tangent to conic (1.1) at the point (f, g, h) has equation (1 + k)(hy + gz) + (hx + fz) + (1 – k)(gx + fy).
(1.4)
The tangents at A, B, C, D therefore have equations: tA: (1 – k)y + z = 0; tB: (1 – k)x + (1 + k)z = 0; tC: x + (1 + k)y = 0; tD: (1 + k)x + 4y + (1 – k)z = 0. (1.5) We can now determine the co-ordinates of the vertices of PQRS which are: P = tA^tB; (k + 1, 1, k – 1); Q = tB^tC: (– (1 + k), 1, 1 – k); R = tC^tD: (k + 1, – 1, k + 3); S = tD^tA: (k – 3, 1, k – 1); U = tA^tC: (k + 1, – 1, 1 – k); T = tB^tD: (3 – (1 + k), k, 1 – k). (1.6) The polar line FGTU has equation x + 2y + z = 0.
(1.7)
The co-ordinates of the points of intersection of the sides of ABCD and PQRS are as follows: 31 = PQ^CD: (2(k + 1), – (1 + k), 2(k – 1)); 24 = SP^BC: (0, 1, k – 1); 34 = SP^CD: (– 2, 1, k – 1); 12 = QR^AB: (k + 1, – 1, 0); 42 = QR^AD: (k + 1, – 1, 2); 13 = RS^AB: (– 4, k + 1, 0); 23 = RS^BC: (0, k – 1, 4); 41 = PQ^AD: (2(k + 1), 1 – k, 2(k – 1)). 2. The 8 point Conic It turns out that the 8 points just listed lie on a conic with an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, where 2
(2.1)
u = v = w = 2(1 – k2), f = (k + 1)(k2 – 2k + 5), g = k2 + 3, h = (1 – k)(k2 + 2k + 5).
(2.2)
3. Lines through E = AC^BD E = AC^BD has co-ordinates (1, 0, 1). It may now be shown that the following 6 lines pass through E: PR: x – (2y + z) = 0; (3.1) QS: x + 4y – z = 0; (3.2) 41 23: (1 – k)x – 4y – (1 – k)z = 0; (3.3) 24 42: x + (k – 1)y – z\ = 0; (3.4) 12 34: x + (1 + k)y – z = 0; (3.5) 31 13: (k + 1)x + 4y – (k + 1)z = 0. (3.6) 4. Four 6 point Conics Altogether we have defined 21 points and in a configuration such as this we must expect a fair number of 6 point conics. Initially we shall concentrate on those involving 41, 23, 24, 42, 12, 34, 31, 13, T, U. The first is the conic 12 31 13 34 T U which has an equation of the form (2.1) where u = w = 2(k + 1), v = 8(k + 1), f = (k + 1)(3 – k), g = 2, h = k2 + 2k + 5.
(4.1)
The second is the conic 24 41 23 42 T U which has an equation also of the form (2.1), where u = w = 2(1 – k), v = 8(1 – k), f = k2 – 2k + 5, g = 2, h = (1 – k)(k + 3). (4.2) The third is 24 12 31 23 T U which has an equation of the form (2.1), where u = 2(1 – k)(k + 3), v = 8(1 – k2), w = 2(1 – k2), f = (k + 1)(k2 – 2k + 5), g = 4, h = (1 – k)(k2 + 4k + 7).
(4.3)
The fourth is 13 34 41 42 T U which again has an equation of the form (2.1), where u = 2(1 – k2)v = 8(1 – k)(k + 3), w = 2(1 – k)(k + 3), f = k3 – k2 – 5k + 13, g = 4, h = (1 – k)(k2 + 4k + 7).
(4.4)
5. The singular points X, Y A curious property of the above lines and conics is that the line PR with equation (3.1), the second of the above conics with equation (4.2) and the fourth of the above conics with equation (4.4) all pass through the same two points X and Y (see the figure). Their co-ordinates satisfy x = 2y + z and y/z = (1/(k3 + 7k2 + 7k – 15)) √(4 – k2 – k)(2√(4 – k – k2) +/- (k + 1)3/2SIGN( k + 1)). (5.1) 3
It would appear that X and Y become complex for certain values of k. The properties of X and Y are rather remarkable, for as can be seen from the figure pairs of lines through X and Y meet on the polar line FGTU. For example XA and YC meet in this way. There are altogether six such pairs of lines. Also there are (according to Cabri II plus) numerous conics passing through X and Y. A few of these are outlined in the figure, one such being X Y S Q C 34. Also there are many pairs of lines through X and Y that lie on the other diagonal QS one such being XD and YA. It would appear that when X and Y exist they are of equal importance to P Q R S F G T U. As far as I am concerned the existence of such points with properties suggested by CABRI II plus is unexpected and the different properties of the diagonal lines PR and QS is also a surprise, considering that the initial configuration does not suggest asymmetric properties.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
4
2
Article: CJB/231/2012
A Quadrilateral and resulting Conics Part 2 A Special Case showing the role of the New Points X and Y Christopher Bradley
To 2
9
T
G
11
U 12
13 To 7 To F
41 X
A
1 34 14 S 5 6 E
P
D
8 To 10
42 Y
R
13 4 24
B
3
Q To 12
C
23
To 31
Figure 1
Abstract: In Article 230 we investigated properties of a configuration of a quadrilateral ABCD inscribed in an ellipse and the common points of tangents to the ellipse at the vertices A, B, C, D. It turns out that there are two remarkable points X and Y through which many lines may be drawn and also through which many conics pass. The analysis of Part 1 was too complicated to illustrate the results, so in this Article we give a numerical presentation that illustrates the properties of the point X and Y, for which the results all hold in the general case. It seems unlikely the general case can be treated analytically, but possibly a pure approach would succeed as the points are members of a Desargues’ involution. Figure 1 above illustrates the line properties of X and Y and Figure 2 below illustrates their Conical properties.
1
To V
T
G
Z
U 13 41
X
A
34 S
W
D 42
P 24
R
E
B
L
C
Y
23
Q
To 31 To 12 1. Introduction, the initial points and lines We use areal co-ordinates throughout with ABC as triangle of reference and D the point with coordinates D(3, – 2, 6). We take the circumscribing ellipse to be the Steiner circum-ellipse with equation 31 yz + zx + xy = 0. (1.1) 2
The equations of BC, CA, AB are, of course x = 0, y = 0, z = 0. It is easy to show that those of AD, BD, CD are respectively 3y + z = 0, 2x – z = 0, 3y + 2x = 0. (1.2) We can now calculate the co-ordinates of the diagonal points of ABCD and these are E = BD^AC: x = 1, y = 0, z = 2, (1.3) F = AD^BC: x = 0, y = – 1, z = 3, (1.4) G = AB^CD: x = 3, y = – 2, z = 0. (1.5) The tangent at (f, g, h) to the ellipse with equation (1.1) is (hy + gz) + (hx + fz) + (gx + fy) = 0.
(1.6)
Therefore the tangent at A has equation y + z = 0, the tangent at B has equation z + x = 0, the tangent at C has equation x + y = 0, and the tangent at D has equation 4x + 9y + z = 0.
(1.7) (1.8) (1.9) (1.10)
We can now compute the co-ordinates of the second quadrilateral PQRS. These are P = tA^tB: x = 1, y = 1. z = – 1, (1.11) Q = tB^tC: x = – 1, y = 1, z = 1, (1.12) R = tC^tD: x = 1, y = – 1, z = 5, (1.13) S = tD^tA: x = 2, y = – 1, z = 1. (1.14) The two external diagonal points are U = tA^tC: : x = 1, y = – 1, z = 1, T = tB^tD: x = – 3, y = 1, z = 3.
(1.15) (1.16)
It transpires that the external diagonal lines of the quadrilaterals ABCD and PQRS are identical. That is F, G, T, U are collinear on the line with equation 2x + 3y + z = 0. (1.17) The sides of the quadrilateral PQRS have equations as follows: PQ: z + x = 0, QR: x + y = 0, RS: 4x + 9y + z = 0, SP: y + z = 0. This is because they are identical to the tangents at A, B, C, D. 2. The eight intersections of the sides of the two quadrilaterals 3
(1.18) (1.19) (1.20) (1.21)
These intersections are key points in the configuration we have just defined in Section 1. In no particular order their co-ordinates are: 31 = PQ^CD: x = – 3, y = 2, z = 3, (2.1) 24 = SP^BC: x = 0, y = – 1, z = 1, (2.2) 34 = SP^CD: x = 3, y = – 2, z = 2, (2.3) 12 = QR^AB: x = 1, y = – 1, z = 0, (2.4) 42 = QR^DA: x = 1, y = – 1, z = 3, (2.5) 13 = RS^AB: x = 9, y = – 4, z = 0, (2.6) 23 = RS^BC: x = 0, y = – 1, z = 9, (2.7) 41 = PQ^DA: x = 3, y = 1, z = – 3. (2.8) The notation is such that PQ, QR, RS, SP carry numbers 1, 2, 3, 4 as do AB, BC, CD, DA. It will be observed that as a consequence of the choice of initial ellipse and the co-ordinates of D, the two points 24 and 12 lie on the line at infinity. This in no way invalidates any of the results that follow. Conics containing these points are such that they pass through the line at infinity in one or other (or both) of these points. In the figures, however, we have chosen a case in which 24 and 12 are finite points. The remarkable property of these 8 points is that they all lie on the same conic. This conic has equation ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, where u = 4, v = 9, w = 1, f = 5, g = 7/2, h = 13/2.
(2.9) (2.10)
3. Lines through E It is no surprise that six further lines in the figure pass through E = AC^BD with co-ordinates E(1, 0, 2). These are as follows: PR: 2x – 3y – z = 0, (3.1) QS: 2x + 3y – z = 0, (3.2) 41 23: 2x – 9y – z = 0, (3.3) 24 42: 2x – y – z = 0, (3.4) 12 34: 2x + 2y – z = 0, (3.5) 31 13: 4x + 9y – 2z = 0. (3.6) 4. Four 6 point conics and the definition of the remarkable points X and Y
4
If one adds the points T, U on the external diagonal line to the eight points defined in Section 2 one can construct four 6 point conics. These are 12 31 13 34 T U: with equation 4x2 + 9y2 + z2 + 4yz + 3zx + 13xy = 0, (4.1) 24 41 23 42 T U with equation 4x2 + 9y2 + z2 + 10yz + 6zx + 10xy = 0, (4.2) 24 12 31 23 T U with equation 10x2 + 9y2 + z2 + 10yz + 9zx + 19xy = 0, (4.3) 13 34 41 42 T U with equation 8x2 + 45y2 + 5z2 + 38yz + 18zx + 38xy = 0. (4.4) The equation of PR is 2x – 3y – z = 0,
(4.5)
It may now be checked that conics (4.3) and (4.4) and PR all meet at points X, Y (and the two conics of course also intersect at T and U so that the four intersections of the two conics are T, U, X, Y. We shall not pursue in this article the involution T↔U , X↔Y). The co-ordinates of X, Y are X: x = (k + 1)/10, y = (k + 6)/15, z = – 1, (4.6) Y: x = (k – 1)/10, y = (k – 6)/15, z = 1. (4.7) In Equations (4.6) and (4.7) k = √6 and in what follows we shall continue to write k for this numerical value. 5. Lines through X and Y intersecting on the diagonal QS Points X and Y have some remarkable properties. In this section we shall show that whenever X is joined by a line through another key point, then there is a line through Y and another key point such that the two lines intersect on the diagonal QS or the line FGTU. Figure 1 shows such lines and the intersection points 1, 2, 3, ... , 14. QS has equation 2x + 3y – z = 0,
(5.1)
2x + 3y + z = 0.
(5.2)
and FGTU has equation
We now give details of these pairs of lines and their points of intersection. Point 1: XU^YT: x = 3k + 3/2, y = – 2k, z = 3.
(5.3)
5
Point 2: XT^YU: x = – 3k + 3/2, y = 2k, z = 3. Point 3: XB^YC: x = – (k + 1)/10, y = (k + 6)/15, z = 1. Point 4: XC^YB: x = (k – 1)/10, y = (6 – k)/15, z = 1. Point 5: XA^YD: x = (k + 11)/10, y = – (k + 6)/15, z = 1. Point 6: XD^YA: x = (11 – k)/10, y = (k – 6)/15, z = 1. Point 7: XC^YA: x = (1 – k)/10, y = (k – 6)/15, z = 1. Point 8: XA^YC: x = (k + 1)/10, y = – (k + 6)/15, z = 1. Point 9: XB^YD: x = – (k + 1)/10, y = (k – 4)/15, z = 1. Point 10: XD^YB: x = (k – 1)/10, y = – (k + 4)/15, z = 1. Point 11: XQ^YS: x = – k/4, y = (k – 2)/6, z = 1. Point 12: XS^YQ: x = k/4, y = – (k + 2)/6, z = 1. Point 13: X 23^ Y 24: x = (k – 2)/2, y = (3 – k)/6, z = 1. Point 14: X 34^Y 13: x = (7 – k)/2, y = (k – 6)/3, z = 1.
(5.4) (5.5) (5.6) (5.7) (5.8) (5.9) (5.10) (5.11) (5.12) (5.13) (5.14) (5.15) (5.16)
It will be observed that the first 12 points are conjugate pairs in which k is replaced by – k in the partner. Points 7, 8, 9, 10, 11, 12 lie on FGTU the others on the diagonal QS. The line XY itself has equation 2x – 3y – z = 0,
(5.17)
and XY passes through F(0, – 1 , 3). See Figure 1 to observe the results of this Section. 6. The conics passing through X and Y We have already defined X and Y as both lying on the two conics 24 12 31 23 T U with equation 10x2 + 9y2 + z2 + 10yz + 9zx + 19xy = 0,
(6.1)
and 13 34 41 42 T U with equation 8x2 + 45y2 + 5z2 + 38yz + 18zx + 38xy = 0.
(6.2)
We now investigate other such conics having an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(6.3)
First we find the conditions that key points lie on the conic (6.3). A: u = 0, B: v = 0, C: w = 0, D: – 24f + 36g – 12h + 9u + 4v + 36w = 0, X: – (2k/15 + 4/5)f – (k/5 +1/5)g + (7k/75 + 4/25)h + (k/50 + 7/100)u + 6
(6.4) (6.5) (6.6) (6.7)
+ (4k/75 + 14/75)v + w = 0. Y: (2k/15 – 4/5)f + (k/5 – 1/5)g + (4/25 – 7k/75)h + (7/100 – k/50)u + + (14/75 – 4k/75)v + w = 0. T: 6f – 18g – 6h + 9u + v + 9w = 0. U: – 2f + 2g – 2h + u + v + w = 0. Q: 2f – 2g – 2h + u + v + w = 0. S: – 2f + 4g – 4h + 4u + v + w = 0. 24: – 2f + v + w = 0. 23: – 18f + v + 81w = 0. 13: – 72h + 81u + 16v = 0. 34: – 8f + 12g – 12h + 9u + 4v + 4w = 0. 41: – 6f – 18g + 6h + 9u + v + 9w = 0. 42: – 6f + 6g – 2h + u + v + 9w = 0. G: – 12h + 9u + 4v = 0.
(6.8) (6.9) (6.10) (6.11) (6.12) (6.13) (6.14) (6.15) (6.16) (6.17) (6.18) (6.19) (6.20)
We now enumerate and give the equations of numerous six or eight point conics, as shown in Figure 2. Conic XYABCD: u = v = w = 0, f = 1, g = 4, h = 10. Conic XYBCTU: u = 28, v = w = 0, f = 4, g = 9, h = 19. Conic XYADTU: u = 0, v = 63, w = 7, f = 25. g = 9, h = 19. Conic XYQSBD: u = 104, v = 0, w = 26, f = 57, g = 20, h = 102. Conic XYACQS: u = 0, v = 78, w = 0, f = 19, g = 24, h = 34. Conic XYTU 41 13 34 42: u = 8, v = 45, w = 5, f = 19, g = 9, h = 19. Conic XYAD 24 23: u = 0, v = 9, f = 5, g = 7, h = 17. Conic XYBC 13 34: u = 16, v = 0, w = 0, f = 3, g = 8, h = 18. Conic XYBC 41 42: u = 4, v = 0, w = 0, f = 2, g = 7, h = 17. Conic TU 41 24 42 23: u = 4, v = 9, w = 1, f = 5, g = 3, h = 5. Conic XYAD 13 34: u = 0, v = 18, w = 2, f = 7, g = 2, h = 4.
(6.21) (6.22) (6.23) (6.24) (6.25) (6.26) (6.27) (6.28) (6.29) (6.30) (6.31)
Cabri II + indicates two further interesting points, which we have called Z and W. These are the intersections of conic XYAD 13 34 (Equation (6.31)) and XYTU 24 12 31 23(Equation (4.3)) With Z also lying on L 42 and W also lying on L41, where L = BC^QS. ZW passes through F as do XY and TU and the pair Z and W also have many conics passing through them. We do not investigate their properties further, but it seems likely that other pairs of points lying on a line through F might be found having systems of conics passing through them. It is one of those cases that as the figure gets larger it then gets even larger. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 7
8
Article: CJB/2012/232
The Coconic Hexagon with Main Diagonals Concurrent Christopher Bradley
Figure 1 Almost complete diagram without labels
1
6c 5b
2f
tF
tB 5
2
6
tC
1e
To 5G
1G
tE To 6c
1
bf
3a
26 tA
15 6b
ac
2a
1f
P
3
A
U 5a
ae
F To ad 13 1d
6e
B
1c
tD
Q 3b bd
C 2d 24
4G 4b
E R D 4c 3e ce
4a 46
T
G
4f
S 5d
To 36
df
35
6d
2e 2G
To 3G
5c
3f
tC To Z
14
be
Figure 2 25 A slightly condensed diagram with labels cf
1. Discussion ABCDEF is a hexagon inscribed in a conic and is such that its main diagonals AD, BE, CF are concurrent at a point G. Sides AB, BC, ... are labelled 1, 2, ... . Points such as 14 are then the intersection of AB and DE. Tangents are drawn at A, B, ..., so that the tangents at A and B meet at P. Point Q is the intersection of the tangents at B and C and so on. The tangents at A, B, … are also labelled tA, tB,… and points such as ac are thus the intersections of tA and tC. It is proved that PQRSTU is a conic and that the six points ac, bf, ae, df, ce, bd lie on a conic. It is akso proved that 13, 26, 15, 46, 35, 24 lie on a conic. Points 14, 25, 36 is the polar line of G with respect to conic ABCDEF and the polar line of PQRSTU, consisting of ad, be, cf, unsurprisingly coincides with it.
2
A point such as 5d is the intersection of line EF and the tangent at D and there are 24 such points and we prove they lie on three 6 point conics, namely 6c 2f 5b 2e 5c 3f and 2a 1e 4a 5d 4b 1d and 1f 1c 2d 4c 4f 5a. The other 6 points lie on the two lines 3 and 6. Finally a point such as 3G means the intersection of line 3 with the appropriate diagonal BGE. It is finally shown that 1G, 3G, 5G are collinear as are 2G, 4G, 6G and that they meet at a point Z on the polar line. 2. Setting the Scene We use homogeneous projective co-ordinates with ABC the triangle of reference and G the unit point. We take the conic in which the hexagon is inscribed to have equation fyz + gzx + hxy = 0. (2.1) The equation of AG is y = z and this meets the conic (1.1) at the point D with co-ordinates D(– f, g + h, g + h). Similarly E and F have co-ordinates E(h + f, – g, h + f) and F(f + g, f + g, – h). The tangent to conic (1.1) at the point (p, q, r) has equation (gr + hq)x + (fr + hp)y + (fq + gp)z = 0.
(2.2)
Giving p, q, r suitable values we find the equations of the tangents to be tA = UP : hy + gz = 0, tB = PQ : fz + hx = 0, tC = QR: gx + fy = 0, tD = RS : (g + h)2x + fgy + hfz = 0, tE = ST : fgx + (h + f)2y + ghz = 0, tF = TU : hfx + ghy + (f + g)2z = 0.
(2.3) (2.4) (2.5) (2.6) (2.7) (2.8)
The intersection points of the tangents have co-ordinates: P(f, g, – h), Q(– f, g, h), R(– f, g, 2g + h), S(– f((h + f – g), – g(g + h – f), 2fg + h(f + g + h)), T(– 2gh –f(f + g + h), g(f + g – h), h(h + f – g)), U(f + 2g, g, – h). (2.9) 3. The conic PQRSTU We may now show that P, Q, R, S, T, U lie on a conic with equation ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, where u = 2fh2(g + h), v = 2fh(f2 + f(g + 2h) + h(g + h)), w = 2f2h(f + g), 3
(3.1)
l = f(f2(g – h) + f(g2 – h2) + gh(g + h)), m = f2(g2 + gh + 2h2) + fg(g + h)2 + g2h(g + h), n = h(f2(g – h) + f(g2 – h2) + gh(g + h)).
(3.2)
4. Points of intersection of tangents The points of intersection of tangents other than P, Q, ... have co-ordinates: ac: (f, – g, h), ad: (f(g – h), – g(g + h), h(g + h)), ae: (f + 2h, – g, h), bd: (– f, g + 2h, h), be: (– f(h + f), g(f – h), h(h + f)), bf: (– f, – 2f – g, h), ce: (f, – g, 2f + h), cf: (f(f + g), – g(f + g), h(g – f)), df: (– f(f + g – h), f(g + 2h) + g(g + h), – h(g + h – f)).
(4.1)
The line ad be cf, the polar of G with respect to conic PQRSTU has the rather satisfying equation (g + h)x + (h + f)y + (f + g)z = 0. (4.2) It may now be shown that the points ac, ae, bd, bf, ce, df lie on a conic with an equation of the form (3.1), where u = 2fgh(g + h), v = – 2fgh(h + f), w = 2fgh(f + g), l = – f(f2(g + h) + f(g + h)2 + gh(g – h)), m = – g(f(f + g + h)(g + 3h) + gh(g + h)), n = h(f2(g – h) – (fg + gh + hf)(g + h)). (4.3) 5. Equations of the sides of ABCDEF and their points of intersection The equation of 1 = AB is z = 0 and the equation of 2 = BC is x = 0. The equation of CD is (g + h)x + fy = 0 and the equation of 4 = DE is (g + h)x + (h + f)y – hz = 0. The equation of 5 = EF is fx – (h + f)y – (f + g)z = 0 and that of 6 = FA is hy + (f + g)z = 0 (5.1) Various points of intersection have co-ordinates as follows: 14: (– (h + f), g + h, 0), 25(0, – (f + g), h + f), 36(f(f + g), – (f + g)(g + h), h(g + h)).
(5.2)
It may be seen that these three points also lie on the common polar line with equation (4.2). Other points of intersection have co-ordinates: 13: (– f, g + h, 0), 15: (h + f, f, 0), 24: (0, h, h + f), 26(0, – (f + g), h), 35: (f(f + g), – (f + g)(g + h), f(f + g + h) + h(g + h)), 46: ( f(f + g + h) + h(g + h), – (f + g)(g + h), h(g + h)).
(5.3)
It may now be shown that the points 13, 15, 24, 26, 35, 46 lie on a conic with equation of the form (3.1), with u = 2hf(g + h), v = – 2hf(h + f), w = 2hf(f + g), l = – f(f(f + g + h) + h(g – h)), m = – f(f + g + h)(g + 2h) – gh(g + h), n = h(f(f – g – h) – h(g + h)). (5.4)
4
6. The 24 points of intersection of the sides and the tangents We give the co-ordinates of these points in order of the sides AB, BC, ... . 1c: (– f, g, 0); 1d: (– fg,(g + h)2, 0); 1e: (– (h + f)2, fg, 0); 1f: (– g, f, 0); 2d: (0, – h, g); 2e: (0, – gh, (h + f)2); 2a: (0, – g, h); 3e: (fg, – g(g + h), f (f + 2g + 2h) + h(g + h)); 3f: (– f(f + g)2, (f + g)2(g + h), h(f2 – g(g + h)), 3a: (fg, – g(g + h), h(g + h)); 3b: (– f, g + h, h); 4f: (f2 + fg + gh, – f(g + h) – g2, h(g – f)); 4a: (fg + h(g + h), – g(g + h), h(g + h)); 4b: ( – f(h + f), f(g + h) + h2, h(h + f)); 4c: (f, – g, f – g); 5a: (h – g, – g, h); 5b: (– f(h + f), – f2 – hf – gh, h(h + f)); 5c: (f(f + g), – g(f + g). f2 + fg + gh); 5d: (f(g – h), – fh – g(g + h), fg + h(g + h)); 6b: (f, f + g, – h); 6c: (f(f + g), – g(f + g), gh); 6d: (f(fg + g2 – h2), – (f + g)(g + h)2, h(g + h)2); 6e; (f2 + f(g + 2h) + h(2g + h), – g((f + g), gh). (6.1) – (6.24) 7. The additional six point Conics Conic 6c 2f 5b 2e 5c 3f This has an equation of the form (3.1), where u = 2gh(f(f + g + 2h) + (g + h)2), v = 2gh(h + f)2, w = 2gh(f + g)2, l = f4 + 2f3(g + h) + f2(g2 + 4gh + h2) + 2fgh(g + h) + 2g2h2, m = g(f3 + f2(2g + h) + f(g + h)2 + g2h), n = h(f3 + f2(g + 2h) + f(2g2 + gh + h2) + g2h).
(7.1) (7.2) (7.3) (7.4) (7.5) (7.6)
Conic 2a 1e 4a 5d 4b 1d This has an equation of the form (3.1), where u = 2fgh(g + h)2, v = 2fgh(h + f)2, w = 2fgh(f2 + f(g + 2h) + g2 + gh + h2), l = f(f2(g2 + h2) + fh(2g2 + gh + 2h2) + h2(2g2 + gh + h2)), m = g(f2(g2 + gh + h2) + fh(g + h)2 + h2(g + h)2)), n = h(f2(2g2 + 2gh + h2) + 2fh(g + h)2 + h2(g + h)2)).
(7.7) (7.8) (7.9) (7.10) (7.11) (7.12)
Conic 1f 1c 2d 4c 4f 5a This has an equation of the form (3.1), where u = 2fgh, v = 2fgh, w = 2hf(f + g – h), l = f(fg + g2 – gh + h2), m = – f2h + f(g2 + h2) + g3,
(7.13) (7.14) (7.15) (7.16) (7.17) 5
n = h(f2 + g2).
(7.18)
The remaining 6 points do not lie on a non-degenerate conic as they lie three by three on lines 3 and 6. 8. Two additional three point lines and where they meet Lines AD, BE CF have equations y = z, z = x, x = y respectively. Points have co-ordinates 1G = CF^AB: (1, 1, 0), 2G = AD^BC: (0, 1, 1), 3G = BE^CD: (f, – (g + h), f), 4G = CF^DE: (h, h, f + g + 2h), 5G= AD^EF: (2f + g + h, f, f),6G – BE^FA: (h, – (f + g), h). (8.1) It may now be shown that 1G, 3G, 5G lie on the line with equation f(x – y) – (f + g + h)z = 0,
(8.2)
And also that 2G, 4G, 6G lie on the line with equation (f + g + h)x + h(y – z) = 0/
(8.3)
These two lines meet at the point Z with co-ordinates Z(h(2f + g + h), – f2 – f(2g + h) – (g + h)2, f(f + g + 2h)).
(8.4)
Also Z lies on the polar line (4.2). David Monk has pointed out that these are Pascal lines so presumably Z is a Steiner point.
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
6
Article: CJB/2012/233
The Conic and the Lines that are Created (Part 1) Christopher Bradley 8 10
Abstract: A Conic is defined by 5 points, no three of which are collinear, Pairs of lines joining these points and the tangents to the conic at its defining points create 10 lines and these lines generate 15 other key lines and many subsidiary ones. 4.11 13 14
b 35 24
8 11 To 8 10
4b 14
To 4b
25 35
10 25
5
ab
3a 14 35 24
a
Ten lines ad 14 be be 25 ac 25 ac 13 bd bd 24 ce 5 Pascal lines ce 35 ad and 5 tangents 25 3a 14 13 4b 25 24 5c 13 35 1d 24 14 2e 35
6 5c
ce
cd
69
9
2 b
c
6 14
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de bc
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14 ae
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d 4 11 2e
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A Conic with 5 lies 1, 2, 3, 4, 5 and 5 tangents a, b, c, d, e and the resulting configuration
bd
1
1. Preliminaries We take the conic to have equation y2 = zx and the five points to be A(1, 0, 0), B(0, 0, 1), C(1, 1, 1), D(1, a, a2) and E(a2, a. 1). Admittedly this is not a perfectly arbitrary set of points, but the resulting configuration exhibits only those that appear in the more general situation. Line 1 is AB with equation y = 0. Line 2 is BC with equation x = y. Line 3 is CD with equation ax – (a + 1)y + z = 0. Line 4 is DE with equation ax – (a2 + 1)y + az = 0. Line 5 is EA with equation y = az.
(1.1) (1.2) (1.3) (1.4) (1.5)
2. The five diagonal points and the conic on which they lie As often in this article we use the notation ij to represent the intersection of lines i and j. Thus the co-ordinates of 5 diagonal points are 13: (1, 0, – a). (2.1) 25: (a, a, 1). (2.2) 14: (1, 0, – 1). (2.3) 2 2 35: (a + a – 1, a , a). (2.4) 2 24: (a, a, a – a + 1). (2.5) It is now possible to work out the equation of the conic through these five points. It has the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (2.6) where we find u = 2a3, v = 2(a3 + 3a2 – a + 1), w = 2a2, f = – 2a(a2 + 1), g = a2(a + 1), h = – 2a2(a + 1). (2.7) 3. The tangents at A, B, C, D, E and the conic through points of intersection of these tangents The tangent to ABCDE at the point (f, g, h) has equation 2gy = fz + hx.
(3.1)
Giving f, g, h appropriate values we find the tangent at A is the line a with equation z = 0, B is the line b with equation x = 0, C is the line c with equation 2y = x + z, D is the line d with equation 2ay = a2x + z,
(3.2) (3.3) (3.4) (3.5)
2
E is the line e with equation 2ay = x + a2z.
(3.6)
The co-ordinates of the points of successive tangents are ab: (0, 1, 0), bc: (0, 1, 2), cd: (2, a + 1, 2a), de: (2a, a2 + 1, 2a), ae: (2a, 1, 0)
(3.7) (3.8) (3.9) (3.10) (3.11)
It is now possible to work out the conic through these five points. Its equation has the form (2.6), where we find: u = – 2a, v = 0, w = 2a(a + 1), f = – 2a(a + 1), g = 1, h = 2a2. (3.12) The intersection points of diagonal lines are: ac: (2, 1, 0); bd: (0, 1, 2a); ce: (2a , a + 1, 2); ad(2, a, 0); (be: ( 0. a, 2).
(3.13)
The conic through these five points has an equation of the form (2.6), where we find u = 2a3, v = 8a2, w = 2a2, f = – 2a(a2 + 1), g = 2a3 – a + 1, h = – 2a2(a + 1).
(3.14)
4. The five Pascal lines As David Monk has pointed out (private communication) the following five lines we obtain are Pascal lines. Line 7: 25 24 3a with equation a(x + z) = (a + 1)y. (4.1) 2 Line 8: 13 25 4b with equation a(ax + z) = (1 + a )y. (4.2) 2 Line 6: 24 13 5c with equation a(ax + z) = (2a – a + 1)y. (4.3) 2 3 2 Line 10: 35 24 1d with equation a(a x + z) = (a + a – a + 1)y. (4.4) 2 2 Line 9: 14 35 2e with equation a (x + z) = (a + 2a – 1)y. (4.5) In the diagram these five lines are drawn as blue dotted lines. 5. The five exterior diagonals of four quadrilaterals formed by missing out one vertex Line 14: ad be 14 with equation a(x + z) = 2y. Line 25: be ac 25 with equation (x + az) = 2y. Line 13: ac bd 13 with equation (ax + z) = 2ay. Line 24: bd ce 24 with equation (a2 + a – 1)x + az = 2a2y. Line 35: ce ad 35 with equation ax + (1 + a – a2)z = 2y.
3
(5.1) (5.2) (5.3) (5.4) (5.5)
In the diagram these five lines are drawn as dark green dotted lines. It may be checked that these five lines are tangent to the conic 14 25 13 24 35 at these very vertices. These points are also intersections of pairs of Pascal lines (blue dotted lines) 6. Other pairs of intersection of Pascal lines (the blue dotted lines) Lines 6 and 7 meet at 67 with co-ordinates (2a, a, 1 – a). Lines 6 and 9 meet at 69 with co-ordinates (2a2 – 1, a2, a(2 – a)). Lines 7 and 10 meet at 7 10 with co-ordinates (a(a + 2), a(a + 1), 1). Lines 8 and 9 meet at 89 with co-ordinates (a2 – 1, a2, 2a) Lines 8 and 10 meet at 8 10 with co-ordinates (a + 1, a, 1 – a).
(6.1) (6.2) (6.3) (6.4) (6.5)
7. Lines passing through four intersections of pairs of Pascal lines Such collinearities are well known in the theory of Pascal lines (Cayley lines). The lines 7 and 24 meet at the point 7 24 with co-ordinates (a(2a + 1), a(a + 1), 1 + a – a2). The line through the four intersections 25 3a 14 7 10 67 has equation a(x + z) = (a + 1)y (7.1) and this line also contains the point 7 24. This is typical behaviour of lines passing through four intersection of pairs of Pascal lines (in the diagram where two blue dotted lines meet) – that is the liner contains the intersection of one line of Section 5 and one line of Section 6. Another example of this behaviour is the line 5 13 5c 24 69 57 6 14 which has equation a(ax + z) = (2a2 – a + 1)y = 0. Here 6.14 has co-ordinates (2a + 1, a, 1 – 2a).
(7.2)
Another example is 13 4b 25 89 8 10 8 11 with equation a(ax + z) = (1 + a2)y with 8 11 having co-ordinates (2a2 – 1, a2, a(2 – a)).
(7.3)
There are five such Pascal lines in all, of which we have given only three. 8. Other lines 4
There are other lines in the diagram whose equations we have not obtained, but Cabri II plus clearly identifies them. These are Line 11 containing 69 7 10 and 8 11. Then the green lines in the figure which are (1) 13 14 4b 24 25 67. (2) 89 10 25 67, (3) 67 14 ac, (4) 24 35 89 13 14, (5) 8 10 25 35 bd, (6) 69 7 24 8 10. It remains a puzzle to me how one can start with a configuration that is unbiased towards 1, 2, 3, 4, 5 and a, b, c, d, e and one can end up with such biased inconsistencies. In Part 2 we determine several conics passing through points defined here in Part 1.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
5
Article: CJB/2012/234
A Conic generates five 8-point Conics (Part 2) Christopher Bradley Abstract: A conic ABCDE with sides 1 = AB, 2 = BC etc. and its tangents a, b, ... at the vertices AB, ... produce 20 intersections through which five 8-point Conics may be drawn.
4 3a d
4b 1d 2e
3 e
be
41
ad
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A
E
5b 5c
5
52 2a ac
5d D
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1c
3b 4c a
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To ce
2d b
1 2 To bd
24
Figure bd 1. Preliminaries We use homogeneous projective co-ordinates throughout and take the conic to have equation y2 = zx. (1.1) 1
We take a general set of points with co-ordinates A(1, 0, 0), B(0, 0, 1), C(1, 1, 1), D(1, a, a2) and E(1, b, b2). The first five lines are now Line 1: AB y = 0, Line 2: BC x = y, Line 3: CD ax + z = (a + 1)y, Line 4: DE abx + z = (a + b)y, Line 5: EA z = by.
(1.2) (1.3) (1.4) (1.5) (1.6)
The five points of intersection (where ij means Line i ^ Line j) other than the vertices have coordinates: 13: (1 , 0, – a), (1.7) 25: (1, 1, b), (1.8) 35: (a – b + 1, a, ab), (1.9) 14: (1, 0, – ab), (1.10) 24: (1, 1, b – ab + a). (1.11) 2. The tangent lines The tangent at (f, g, h) to the conic y2 = zx is 2gy = hx + fz. Using the co-ordinates of the vertices in Section 1 we find the equations of the tangents at the vertices are: Line a: z = 0, (2.1) Line b: x = 0, (2.2) Line c: 2y = x + z, (2.3) 2 Line d: 2ay = a x + z, (2.4) 2 Line e: 2by = b x + z. (2.5) 3. Intersection of lines between lines of Section 1 and those of Section 2 We give the co-ordinates of the requisite points. These are: 1e: (1, 0, – b2), 4a: (a + b, ab, 0), 5d: (2a – b, a2, a2b), 3e: (2b – 1 – a, b2 – a, b(ab – 2a + b)), 1d: (1, 0, – a2), 3a: (a + 1, a, 0), 4b: (0, 1, a + b), 2e: (1, 1, b(2 – b)), 5b: (0, 1, b), 2
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9)
2a: (1, 1, 0), 5c: (2 – b, 1, b), 1c: (1, 0, – 1), 3b: (0, 1, a + 1), 2d: (1, 1, a(2 – a)), 4c: (a + b – 2, ab – 1, 2ab – a – b).
(3.10) (3.11) (3.12) (3.13) (3.14) (3.15)
4. The Five 8-point Conics It is rather remarkable that the twenty points listed in Sections 1 and 3 produce five 8-point conics with 2 conics passing through each point. We give the equations of these conics in the standard form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, (4.1) by displaying the values of the constants u, v, w, f, g, h in each case. CONIC 1 passing through 14 13 2e 5c 2d 4c 3e 5d u = 2a2b, v = – 2b(a2 + ab – 7a + b), w = 2, f = a2 – ab – 3a + b2 – 2b, g = a(b + 1), h = b(a2(b – 2) – a(b + 3) + b).
(4.2)
CONIC 2 passing through 25 24 1e 4a 5d 3e 1d 3a u = 2a2b2, v = 2b(a + 1)(a + b), w = 2, f = – a2 + a(b – 1) – b(b + 2), g = a2 + b2, h = – ab(ab + a + 2b).
(4.3)
CONIC 3 passing through 13 35 4b 2e 5b 2a 4a 1e u = 2ab2, v = 2b(a + b), w = 2, f = – a – 2b, g = a + b2, h = – b(ab + a + b).
(4.4)
CONIC 4 passing through 24 14 5c 1c 3b 3a 5b 2a u = 2ab, v = 2b(a + 1), w = 2, f = – a – b – 1, g = ab + 1, h = – b(2a + 1).
(4.5)
CONIC 5 passing through 35 25 2d 4c 1d 4b 1c 3b u = 2a2, v = 2(a + 1)(a + b), w = 2, f = – 2a – b – 1, g = 1 + a2, h = – a2(b + 2) + a(b – 1) – b.
(4.6)
There are some additional features demonstrated by Cabri II plus, but we do not provide the analysis. Denoting intersections of tangent lines a and c as ac etc., we find the following five lines: ac B and the unnamed intersections of Conics 1 and 2, bd C and the unnamed intersections of Conics 2 and 3, 3
ce D and the unnamed intersections of conics 3 and 4, da E and the unnamed intersections of Conics 4 and 5, eb A and the unnamed intersections of Conics 5 and 1.
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4
Article: CJB/235/2012
The Truth and the whole Truth about a Quadrilateral Christopher Bradley Abstract: A quadrilateral is inscribed in a conic and its sides and diagonals and the tangents at its vertices are drawn. A study is made of the lines and conics that pass through the points of intersection. In all a total of eighteen conics emerge three of which pass through eight points and fifteen through six points. 31
d
1
3
a6
S ac
P
5
c6
d1 b4
2
a3
da A
D c4 U cd d2 C
b 56 ab B
bc
a
4 42
To b5 b3 d5
T
6
c
a2
bd
c1 Q
R
Figure 1 The points and lines generated by ABCD and the tangents at its vertices 1. The sides and diagonal points of quadrilateral ABCD We use homogeneous projective co-ordinates throughout and we take the conic in which ABCD is inscribed to have equation 1
b5
yz + zx + xy = 0.
(1.1)
Let ABC be the triangle of reference and suppose D has co-ordinates (x, y, z), where x = – t(1 – t), y = 1 – t, z = t. (1.2) The sides and diagonals have the following equations: Line 1: AB, z = 0, Line 2: BC, x = 0, Line 3: CD, x + ty = 0, Line 4: DA, (1 – t)z = ty, Line 5: AC, y = 0, Line 6: BD, x + (1 – t)z = 0.
(1.3) (1.4) (1.5) (1.6) (1.7) (1.8)
We use the notation that Line X and line Y meet at the point xy. The diagonal points have coordinates: Point 13, (– t, 1, 0), (1.9) Point 42, (0, 1 – t, t), (1.10) Point 56, ((1 – t), 0, - 1) (1.11) 2. The tangents The tangents at point (f, g, h) to conic (1.1) has equation (g + h)x + (h + f)y + (f + g)z = 0.
(2.1)
We call the tangents to conic (1.1) at A, B, C, D lines a, b, c, d. These lines have equations Line a: y + z = 0, (2.2) Line b: z + x = 0, (2.3) Line c: x + y = 0, (2.4) 2 2 Line d: x + t y + (1 – t) z = 0. (2.5) Points created by intersections of pairs of tangents have co-ordinates: ab : (1, 1, –1), ac : (1, – 1, 1), ad : (2t – 1, – 1, 1), bc: (– 1, 1, 1), bd ; (– t, 2 – t, t), cd : (t – 1, 1 – t, 1 + t).
2
(2.6) (2.7) (2.8) (2.9) (2.10) (2.11)
3. Intersection of quadrilateral lines and tangents at their vertices The co-ordinates of these points are now recorded: a2 : (0, 1. – 1), a3 : (t, – 1, 1), b3 : (– t, 1, t), b4 : (– t, 1 – t, t), c4 : (t – 1, 1 – t, t), c1 : (1, – 1, 0), d2 : (0, – (t – 1)2, t2), d1 : (– t2, 1, 0), a6 : (1 – t, 1, – 1), b5 : (1, 0, – 1), c6 : (t – 1, 1 – t, 1), d5 : (– (t – 1)2, 0, 1).
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9) (3.10) (3.11) (3.12)
4. Four major lines and six special points It turns out that there are four very important lines that may be identified from these points. These are, together with their equations, Line 7: a2 b5 c1, x + y + z = 0, (4.1) 2 2 2 Line 8: b3 c6 d2, (t – t + 1)x + t y + (t – 1) z = 0, (4.2) 2 2 Line 9 : a3 c4 d5, x + (t – t + 1)y + (t – 1) z = 0, (4.3) Line 10 : d1 b4 a6, x + t2y + (t2 – t + 1) = 0. (4.4) The reason for their importance is that these lines create a quadrilateral PQRS with diagonal points T, U and curiously 31. In other words these two quadrilaterals share one diagonal point. We now describe where the six points P, Q, R, S, T, U are located and give their co-ordinates. See also the figure for the visual treat of this quadrilateral associated with ABCD. P = c4 d5 a3 ^ d1 a6 b4 : (– 2t2 + 2t – 1, t, 1 – t), Q = a2 b5 c1 ^ b3 c6 d2 : (1 – 2t, t, t – 1), R = a2 b5 c1 ^ c4 d5 a3 : (1, t – 2, 1 – t), S = b3 c6 d2 ^ d1 a6 b4 : (– t, t2 – 2t + 2), T = a2 b5 c1 ^ d1 a6 b4 : (– 1, – t, t + 1), U = b3 c6 d2 ^ c4 d5 a3 : (t – 1, t(1 – t), t2 + 1).
(4.5) (4.6) (4.7) (4.8) (4.9) (4.10)
We now give the equations of the sides of the quadrilateral PQRS and name the points that lie on them, 3
PQ: x + ty + (1 – t)z = 0. Points 31 ad 56 bc lie on this line. RS : x + ty + (t – 1)z = 0. Points 31 ac 42 bd lie on this line. TU: x – ty + (1 – t)z = 0. Points ab 56 cd 42 lie on this line.
(4.11) (4.12) (4.13)
QR is Line 7 which also contains points T a2 c1 b5. QS is Line 8 which also contains U b3 d2 c6. PR is Line 9 which also contains U a3 c4 d5. PS is Line 10 which also contains T b4 d1 a6. 5. Four more lines through 31 and four more lines through 42 and six more lines through 56 These eight lines have equations as follows: 31 a6 b5 : x + ty + (t – 1)2z = 0, 31 c6 d5 : x + ty + z = 0, 31 a2 b4 : x + ty + tz = 0, 31 c4 d2 : tx + t2y + (t – 1)2z = 0,
(5.1) (5.2) (5.3) (5.4)
42 c1 b3 : tx + ty + (t – 1)z = 0, 42 a3 d1 : x + t2y + t(t – 1)z = 0, 42 a6 d5 : x + t(t – 1)y + (t – 1)2z = 0, 42 c6 b5 : (t – 1)x + ty + (t – 1)z = 0.
(5.5) (5,6) (5.7) (5.8)
56 b4 d2 : (t – 1)x – t2y – (t – 1)2z = 0, 56 d1 b3 : x + t2y + (1 – t)z = 0, 56 a2 c4: x + (1 – t)y + (1 – t)z = 0, 56 a3 c1: x + y – (t – 1)z = 0, 56 b5 d5: y = 0, 56 a6 c6 : x + (1 – t)z = 0.
(5.9) (5.10) (5.11) (5.12) (5.13) (5.14)
6. Conics passing through six of the points A, B, C, D, P, Q, R, S, T, U We take conics to be of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(6.1)
and specify each conic by its values of u, v, w, f, g, h.# Conic ABCDPQ : u = v = w = 0, f = 4t3 – 6t2 + 4t – 1, g = t3, h = (t – 1)3. (6.2) 3 2 3 2 Conic ABCDRS : u = v = w = 0, f = – 1, g = t – 4t + 6t – 4, h = t – 3t + 3t – 1. (6.3) Conic ABPQRS : u = v = 0, f = 2t2 – t + 1, g = t2 – t + 2, h = (t – 1)2. (6.4) 2 2 2 4 3 2 Conic CDPQRS : w = 0, u = 2(2t – 3t + 2), v = 2t (2t – 3t + 2), f = 2t – 5t + 5t – 3t + 1, g = t4 – 3t3 + 5t2 – 5t + 2, h = t4 + 1. (6.5) 4
31
d
1
S
3
a6
ac
P 5
c6
a3
d1
2
da
b b4
A
D c4 U cd
4
42
d2
56
C
ab B
bc
a
b3 d5
T
c
6 a2
bd c1
Q R
b5
Figure 2 The eighteen conics Conic PQTUBD : u = 2(2t2 – t + 1), v = 0, w = 2(t – 1)2(2t2 – t + 1), f = t2(2t2 – 3t + 2), g = t4 – 4t3 + 6t2 – 4t + 2, h = t2(t2 – t + 2). (6 .6) 2 2 2 2 Conic PQTUAC : u = w = 0, v = 2(2t – t + 1), f = 2t – 3t + 2, g = t , h = t – t + 2. (6.7) 2 2 2 2 4 3 2 Conic RSTUAD: u = 0, v = 2t (t – t + 2), w = 2(t – 1) (t – t + 2), f = 2t – 4t + 6t – 4t + 1, g = 2t2 – 3t + 2, h = 2t2 – t + 1. (6.8) 2 2 2 Conic RSTUBC: v = w = 0, u = 2(t – t + 2), f = 1, g = 2t – 3t + 2, h = 2t – t + 1. (6.9) 7. Six more six-point conics It is difficult to be sure that we have located all the six-point conics. Here are the equations of those we have found. Conic P Q b5 d5 a6 c6: u = 2t, v = 2t3, w = 2t(t – 1)2, f = 2t3 – 4t2 + 4t – 1, g = t(t2 – 2t + 2), h = t3 – t2 + 3t – 1. (7.1)
5
Conic R S b5 d5 a6 c6: u = 2(t2 – 2t + 2), v = 2t2(t2 – 2t + 2), w = 2(t – 1)2(t2 – 2t + 2), f = 2t4 – 6t3 + 8t2 – 5t + 2, g = t4 – 4t3 + 8t2 – 8t + 4, h = t4 – 3t3 + 5t2 – 3t + 2. Conic T U b5 d5 a6 c6 : u = 2t, v = 2t3, w = 2t(t – 1)2, f = 2t3 – 2t2 + 1, g = t(t2 – 2t + 2), h = t3 – t2 + t + 1. Conic ac bd b5 d5 a6 c6: u = 2, v = 2t2 w = 2(t – 1)2, f = 2t2 – 3t + 2, g = t2 – 2t + 2, h = t2 – t + 2. Conic ab cd b5 d5 a6 c6: u = 2, v = 2t2, w = 2(t – 1)2 f = 2t2 – t, g = t2 – 2t + 2, h = t(t – 1). 2 2 Conic ad bc b5 d5 a6 c6: u = 2, v = 2t , w = 2(t – 1) , f = t, g = t2 – 2t + 2, h = t(t + 1).
(7.2) (7.3) (7.4) (7.5) (7.6)
8. Three eight-point conics These are the most interesting conic as they involve line and tangent intersections and pas through eight points. Conic c1 d1 a3 b3 b5 d5 a6 c6: u = 2, v = 2t2, w = 2(t – 1)2, f = t2 – t + 1, g = t2 – 2t + 2, h = t2 + 1. (8.1) 2 2 2 2 Conic a2 d2 b4 c4 b5 d5 a6 c6: u = 2, v = 2t , w = 2(t – 1) , f = 2t – 2t + 1, g = t – 2t + 2, h = t2 – t + 1. (8.2) 2 2 2 2 Conic c1 d1 b4 c4 a2 d2 b3 a3: u = 2, v = 2t , w = 2(t – 1) f = 2t – 2t + 1, g = t – t + 1, h = t2 + 1. (8.3) The similarity of many of these equations is striking and it is because many of these conics intersect in pairs at four points lying on two of the lines featured earlier.
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7
Article: CJB/2012/236
A Nice Conic and a Nasty Circle Christopher Bradley Abstract: A triangle ABC and its circumcircle S are given. The line through B parallel to CA and the line through C parallel to AB meet at L. Points M, N are similarly defined by cyclic change. Tangents to S at B and C meet at U, with V, W defined by cyclic change. It is proved that LMNUVW is a conic and the centre of circle UVW is shown to lie on the Euler line of ABC.
W
A
N
M V
Q O H C
B
L
Figure
1
U
1. Introduction We use areal co-ordinates throughout. The equation of the circumcircle S is a2yz + b2zx + c2xy = 0.
(1.1)
The line parallel to BC through A has equation y + z = 0.
(1.2)
The lines parallel to CA through B and parallel to AB through C are z + x = 0 and x + y = 0.
(1.3)
These last two lines meet at L with co-ordinates L(– 1, 1, 1). Similarly M and N have coordinates M(1, – 1, 1) and N(1, 1, – 1). The tangent at A(1, 0, 0) to the circle (1.1) is b2z + c2y = 0.
(1.4)
The tangent at B has equation c2x + a2z = 0,
(1.5)
a2y + b2x = 0.
(1.6)
and the tangent at C has equation
These last two tangents meet at U with co-ordinates U(– a2, b2, c2). Similarly V, W have coordinates V(a2, – b2, c2) and W(a2, b2, – c2). 2. The conic through L, M, N, U, V, W A conic has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(2.1)
Substituting the co-ordinates of the six points L, M, N, U, V, W in equation (2.1) we find equations for the constants u, v, w, f, g, h and these are : f = g = h = 0, u = b4 – c4, v = c4 – a4, w = a4 – b4. 3. The circle UVW A circle has an equation of the form a2yz + b2zx + c2xy + (x + y + z)(ux + vy + wz) = 0.
2
(3.1)
Putting in the co-ordinates of U, V, W we obtain equations for u, v, w and find u = a2b2c2/((c2 + a2 – b2)(a2 + b2 – c2)), v = a2b2c2/(( a2 + b2 – c2)(b2 + c2 – a2 )), w = a2b2c2/(( b2 + c2 – a2)(c2 + a2 – b2 )).
(3.2) (3.3) (3.4)
4. The centre of circle UVW Using standard procedures it may be shown that the x-co-ordinate of the centre Q of the circle UVW is x = a2(a14 – 3a12(b2 + c2) + a10(b4 + 6b2c2 + c4) + a8(5b6 – 3b4c2 – 3b2c4 + 5c6) – a6(5b8 – 2b4c4 + 5c8) – a4(b10 – 3b8c2 + 2b6c4 + 2b4c6 – 3b2c8 + c10) + a2(3b12 – 6b10c2 + 5b8c4 – 4b6c6 + 5b4c8 – 6b2c10 + 3c12) – (b2 – c2)2(b4 + c4)*b6 – b4c2 – b2c4 + c6)). (4.1) The y- and z- co-ordinates follow by cyclic change of a, b, c. It may now be checked that the point Q lies on the Euler line whose equation is (b2 – c2)(b2 + c2 – a2)x + (c2 – a2)(c2 + a2 – b2)y + (a2 – b2)(a2 + b2 – c2)z = 0. It is a matter of interest if there is a pure proof that Q lies on the Euler line.
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3
(4.2)
Article: CJB/2012/237
How to Construct the Ex-points of a given Point Christopher Bradley Abstract: It is shown how knowing the Ex-points of one point it is possible to construct the Expoints of other points.
T
A N
M S Q G B
P C
L R
Figure 1. Discussion of the procedure We take the Ex-points of the centroid G(1, 1, 1) which are the points L(– 1, 1, 1), M(1, – 1, 1), N(1, 1, – 1). These may be constructed by drawing lines through A, B, C parallel to BC, CA, AB respectively. See the figure. We now place the point P (p, q, r) in the diagram whose Ex-points 1
we wish to construct. The next step is to draw the conic G, L, M, N, P and one finds that it contains the Ex-points of P. The Ex-points R, S, T may then be determined as the points of intersection of AP, BP, CP with the conic. 2. The analysis The conic passing through G, L, M, N, P has equation (q2 – r2)x2 + (r2 – p2)y2 + (p2 – q2)z2 = 0.
(2.1)
This may easily be checked and also it may be checked that the Ex-points R(– p, q, r), S(p, – q, r), T(p, q, – r) also lie on conic (2.1). It is interesting that the circles BCR, CAS, ABT have a common point Q with co-ordinates x = p(p – q – r)(a2qr + b2rp – c2pq)(a2qr – b2rp + c2pq), (2.2) with y and z co-ordinates following by cyclic change of a, b, c and p, q, r. We call this the associate of P. Unfortunately this mapping is not an involution. It may be checked that the associate of G is the orthocentre H, the associate of I is I itself, and the associate of the symmedian point K is the circumcentre O. 3. The generalization The conic with equation (m2r2 – n2q2)x2 + (n2p2 – l2r2)y2 + (l2q2 – m2p2)z2 = 0
(3.1)
passes through the points (l, m, n) and (p, q, r) and the three Ex-centres of each of these points. The set of all such conics are eight-point conics of some significance.
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2
Article: CJB/238/2012
When Two Quadrilaterals are in Complete Perspective Christopher Bradley Abstract: If two quadrilaterals ABCD and PQRS are such that AP, BQ, CR, DS are concurrent at X then S may be moved into one and only one new position on DX so that the four Desargues’ axes coincide.
cd
ab
bd ad
S D
P
X
A
C
B
bc
R Q
ac
ABCD and PQRS in Complete Perspective
1
1. Summary If two quadrilaterals ABCD and PQRS are such that AP, BQ, CR, DS are concurrent at a point X, then there are four Desargues’ axes of perspective corresponding to the four triangles ABC, ABD, ACD, BCD. It is shown in this article that S may be moved into a new position so that the four Desargues’ axes coincide and that the new position of S is unique. We then say that the quadrilaterals are in complete perspective. 2. Positions of the vertices of the quadrilateral We use projective co-ordinates throughout with ABC as triangle of reference and D the point with co-ordinates D(f, g, h). We suppose that the point X of concurrence of AP, BQ, CR, DS has co-ordinates X(1, 1, 1). The there exist constants s, k, m, n such that the co-ordinates of P, Q, R, S are P(1 + s, s, s), Q(k, 1 + k, k), R(m, m, 1 + m), S(f + n, g + n, h + n). 3. The equations of the sides and diagonals of the quadrilaterals With the co-ordinates as specified in Section 2 the equations of the sides and diagonals of the two quadrilaterals are: AB: z = 0, (3.1) BC: x = 0, (3.2) CD: fy = gx, (3.3) DA: gz = hy, (3.4) AC: y = 0, (3.5) BD: fz = hx, (3.6) PQ: (k + s + 1)z = sx + ky, (3.7) QR: (k + m + 1)x = ky + mz, (3.8) RS: (g(m + 1) – hm + n)x = (f(m + 1) – hm + n)y + m(g – f)z, (3.9) SP: (h – g)sx + (fs – h(s + 1) – n)y = (fs – g(s + 1) – n)z, (3.10) PR: sx + mz = (m + s + 1)y, (3.11) QS: k(f – h)y = (gk – h(k + 1) – n)x + (f(k + 1) – gk + n)z. (3.12) 4. The points where corresponding sides meet AB^PQ = ab: (– k, s, 0), BC^QR = bc: (0 – m, k), CD^RS = cd: (fm, gm, hm – n), DA^SP = da: (fs – n, gs, hs),
(4.1) (4.2) (4.3) (4.4) 2
AC^PR = ac: (– m, 0, s), BD^QS = bd: (fk, gk – n, hk).
(4.5) (4.6)
5. The four Desargues’ lines of perspective and when they coincide Line ab bc ac: sx + ky + mz = 0, Line ab bd da: hsx + hky = (fs + gk – n)z, Line ac da cd: gsx + gmz = (fs + hm – n)y, Line bc cd bd: fky + fmz = (gk + hm – n)x.
(5.1) (5.2) (5.3) (5.4)
These four Desargues’ lines of perspective coincide if and only if n = fs + gk + hm.
(5.5)
This means that there is one and only one point on DX for the point S such that the four Desargues’ lines of perspective coincide. Thus a construction for two quadrilaterals ABCD and PQRS to be in complete perspective is as follows: Let ABCD be any convex quadrilateral. Choose PQR to be any triangle such that AP, BQ, CR are concurrent at a point X. Now there is just one point on DX where S may be placed so that the four Desargues’ axes coincide and then we have complete perspective.
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Article: CJB/239/2012
More on Complete perspective Christopher Bradley Abstract: It is shown that if two quadrilaterals ABCD and PQRS are such that AP, BQ, CR, DS are concurrent at a point X (and with the notation that AB^PQ = ab etc.) then if ab, bc, cd are collinear the quadrilaterals are in complete perspective (see Article 238).
D A A'
X C
bc
B'
B
da
To AC
bd ab
C' cd
D'
Illustration of complete perspective 1. Continuation of Article 238 We prove now that two quadrilaterals ABCD PQRS with AP, BQ, CR, DS concurrent at a point X, then (with AB^PQ = ab etc.) then provided ab, bc, cd are collinear then the two quadrilaterals are in complete perspective.
The equation of the line ab bc is sx + ky + mz = 0.
(1.1)
The condition that cd lies on this line is n = as + bk + cm
(1.2)
From Article 238 this is the sole condition for the two quadrilaterals to be in complete perspective.
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Article: CJB/240/2012
Pentagons in Complete Perspective Christopher Bradley Abstract: If ABCDE and PQRST are pentagons such that AP, BQ, CR, DS, ET are concurrent at a point X. Using the notation AB^PQ = ab etc. it is proved that if ab, bc, cd, de are collinear then the pentagons are in complete perspective, that is the ten Desargues’ axes merge into the one line ab bc.
ab ad A' A
D'
E
ea de ac
D B
E'
be X cd bc
C' C
B'
To bd ce
Pentagons in Complete Perspective 1. The ten points that are vertices of the two pentagons
1
Projective co-ordinates are used throughout. In the pentagon ABCDE we take ABC to be the triangle of reference, D to have co-ordinates D(a, b, c) and E to have co-ordinates E(d, e, f). We suppose that AP, BQ, CR, DS, ET are concurrent at X(1, 1, 1) and then there exist constants s, k, m, n, t such that the co-ordinates of the second pentagon are P(1 + s, s, s), Q(k, 1 + k, k), R(m, m, 1 + m), S(a + n, b + n, c + n), T(d + t, e + t, f + t). 2. The equations of all the sides and diagonals With the co-ordinates as specified in Section 1 we find the sides and diagonals to have equations as follows: AB: z = 0, (2.1) BC: x = 0, (2.2) CD: ay = bx, (2.3) DE: (bf – ce)x + (cd – af)y + (ae – bd)z = 0, (2.4) AE: fy = ez, (2.5) AC: y = 0, (2.6) AD: cy = bz, (2.7) BD: cx = az, (2.8) BE: fx = dz, (2.9) CE: ex = dy, (2.10) PQ: sx + ky = (s + k + 1)z, (2.11) QR: ky + mz = (k + m + 1)x, (2.12) RS: (b(m + 1) – cm + n)x = (a(m + 1)y – cm + n)y + m(b – a)z, (2.13) ST: (b(f + t) – c(e + t) – n(e – f))x – (a(f + t) – c(d + t) – n(d – f))y + (a(e + t) – b(d + t) – n(d – e))z = 0, (2.14) TP: s(f – e)x + (ds – f(s + 1) – t)y – (ds – e(s + 1) – t)z = 0, (2.15) PR: sx + mz = (m + s + 1)y, (2.16) PS: s(c – b)x + (as – c(s + 1) – n)y – (as – b(s + 1) – n)z = 0, (2.17) QS: k(a – c)y = (bk – c(k + 1) – n)x + (a(k + 1) – bk + n)z, (2.18) QT: k(d – f)y = (ek – f(k + 1) – t)x + (d(k + 1) – ek + t)z, (2.19) RT: m(e – d)z + (d(m + 1) – fm + t)y = (e(m + 1) – fm + t)x. (2.20) 3. The points of intersection of corresponding sides and diagonals The co-ordinates of the 10 points are as follows: AB^PQ = ab (– k, s, 0), BC^QR = bc (0, – m, k), CD^RS = cd (am, bm, cm – n), DE^ST = de (at – dn, bt – en, ct – fn), AE^PT = ae (ds – t, es, fs), 2
(3.1) (3.2) (3.3) (3.4) (3.5)
AC^PR = ac (– m, 0, s), AD^PS = ad (as – n, bs, cs), BD^QS = bd (ak, bk – n, ck), BE^QT = be (dk, ek – t, fk), CE^RT = ce (dm, em, fm – t).
(3.6) (3.7) (3.8) (3.9) (3.10)
4. The conditions that the above 10 points lie on a line We now show that provided cd and de lie on the line ab bc then the other six points of intersection listed in Section 3 also lie on the line. This is a remarkable result, in that only two conditions are necessary for the 10 Desargues’ axes to coincide. The equation of the line ab bc is sx + ky + mz = 0.
(4.1)
The condition that cd lies on this line is as + bk + cm = n.
(4.2)
The condition that de lies on this line (taking account of Equation (4.1)) is ds + ek + fm = t.
(4.3)
What the conditions (4.2) and (4.3) mean in practical terms is as follows: The pentagon ABCDE may be taken arbitrarily. The pentagon PQRST is chosen so that AP, BQ, CR, DS, ET are concurrent. P, Q, R may be taken arbitrarily on the lines AX, BX, CX respectively. But once this is done the positions of S, T on the lines DX, EX respectively are unique. Only one position of both S and T is feasible for complete perspective. It may be checked that provided Equations (4.2) and (4.3) hold then all the points ae, ac, ad, bd, be, ce lie on the line with Equation (4.1). (Note that in the Figure P, Q, R, S, T are labelled A', B', C', D', E' respectively.)
Flat 4, Terrill Court, 12-14 Apsley Road, BRISTOL BS8 2SP.
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Article: CJB/241/2012
Quadrilaterals in Perspective inscribed in a Conic Christopher Bradley Abstract: If ABCD is a quadrilateral inscribed in a conic S and P is any point not on S and AP, BP, CP, DP meet S again in points A', B', C', D' then the quadrilaterals ABCD and A'B'C'D' are in complete perspective. In particular if AC^BD = E and A'C'^B'D' = E' then E, E', P are collinear.
D
A E
B' C
P
C' E' B D'
A'
Figure 1. The vertices of the quadrilaterals We take the conic S to have equation y2 = zx and the points A, B, C to have co-ordinates A(1, 0, 0), B(1, 1, 1), C(0, 0, 1) and D we take to be the general point D(1, t, t2), with t ≠ 0, 1, ∞. Let the vertex of perspective P have co-ordinates (f, g, h). The equation of AP is hy = gz and this meets S at the point A' with co-ordinates A'(g2, gh, h2).
1
The equation of BP is (h – g)x + (f – h)y + (g – f)z = 0,
(1.1)
and this meets S at the point B' with co-ordinates ((f – g)2, (f – g)(g – h). (g – h)2). The equation of CP is gx = fy and this meets S at the point C' with co-ordinates (f2, fg. g2). The equation of DP is t(h – gt)x + (ft2 – h)y + (g – ft)z = 0,
(1.2)
and this meets S at the point D' with co-ordinates ((ft – g )2, (ft – g)(gt – h), (gt – h)2). 2. The points E, E' and their collinearity with P The equation of AC is y = 0 and the equation of BD is tx – (t + 1)y + z = 0.
(2.1)
These lines meet at E with co-ordinates E(– 1, 0, t). The equation of A'C' is ghx – (fh + g2)y + fgz = 0,
(2.2)
(g – h)(gt – h)x + (g(g(t + 1) – 2h) – f(2gt – h(t + 1)))y + (f – g)(ft – g)z = 0,
(2.3)
and the equation of B'D' is
and these two lines meet at the point E' with co-ordinates (f2t – g2), g(ft – h), g2t – h2). The equation of PE is gtx – (ft + h)y + gz = 0, and it may now be checked that E' lies on this line.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
(2.4)
3
Article: CJB/242/2012
Perspective Quadrilaterals Christopher Bradley Abstract: Two quadrilaterals ABCD and A'B'C'D' are such that AA', BB', CC', DD' are concurrent at a point X. It is found that a single condition is such that the four Desargues’ axes of perspective coincide. It is also the case that when this happens the diagonal point triangles EFG and E'F'G' are also in perspective with vertex of perspective X and the same Desargues’ axis of perspective. See also Article 238.
F
D A A' bc
G' G B' eg da fg bd ab
E
E'
X C
B To AC
ef
C' cd F'
D'
Figure
1
1. The vertices of the two quadrilaterals We suppose that ABC is the triangle of reference and we use projective co-ordinates throughout. Take D to have co-ordinates D(a, b, c). Let X be the vertex of perspective with co-ordinates X(1, 1, 1). There then exist constants l, m, n, k so that the co-ordinates of A', B', C', D' are as follows: A'(1 + l, l, l), (1.1) B'(m, 1 + m, m), (1.2) C'(n, n, 1 + n), (1.3) D'(a + k, b + k, c + k). (1.4) 2. Corresponding sides The equation of AB is z = 0. The equation of A'B' is lx + my = (l + m + 1)z.
(2.1)
The equation of BC is x = 0. The equation of B'C' is my + nz = (m + n + 1)x.
(2.2)
Line AB meets line A'B' at ab with co-ordinates (– m, l, 0) and the line BC meets the line B'C' at bc with co-ordinates (0, – n. m). ab and bc lie on the line lx + my + nz = 0 (2.3) and this we take to be the equation of the Desargues’ axis, though one must expect one or more conditions for the meets of other pairs of corresponding sides to lie on this line. The equation of CD is ay = bx and the equation of C'D' is (b(n + 1) – cn + k)x + n(a – b)z = (a(n + 1) – cn + k)y.
(2.4)
Line CD meets line C'D' at cd with co-ordinates (an, bn, cn – k). The condition that cd lies on ab bc is k = al + bm + cn. (2.5) It turns out that condition (2.5) is sufficient for all other pairs of corresponding sides to meet on ab bc. That is the four Desargues’ axes coincide when the parameter k satisfies Equation (2.5). Geometrically this means there is one and only one position for D' on DX for which the four Desargues’ axes coincide. For example ad has co-ordinates (al – k, bl, cl) and (2.5) is the condition for this point to lie on the line ab bc. 3. The diagonal point triangle 2
The equation of AC is y = 0 and the equation of BD is cx = az. The diagonal point E = AC^BD therefore has co-ordinates (a, 0, c). The equation of A'C' is lx + nz = (l + n + 1)y,
(3.1)
and the equation of B'D' is (bm – c(m + 1) – k)x + m(c – a)y + (a(m + 1) – bm + k)z.
(3.2)
These two lines meet at the point E' with co-ordinates (x, y, z), where x = a(l(m + 1) + m + n + 1) – bm(n + l + 1) + cmn + k(n + l + 1), y = al(m + 1) – bm(n + l) + cn(m + 1) + k(n + l), z = alm – bm(n + l + 1) + c(l + (m + 1)(n + 1)) + k(n + l + 1).
(3.3) (3.4) (3.5)
The equation of EP is az + y(c – a) = cx.
(3.6)
It is now found that the condition that E' lies on EP is the condition (2.5). Now AB has equation z = 0 and CD has equation bx = ay. The diagonal point F = AB^CD therefore has co-ordinates (a, b, 0). The line A'B' has equation lx + my = (l + m + 1)z, (3.7) and the equation of C'D' is (b(n + 1) – cn + k) + n(a – b)z = (a(n + 1) – cn + k)y.
(3.8)
The diagonal point F' = A'B'^C'D' has co-ordinates x = a(l(n + 1) + m + n + 1) + bmn – (l + m + 1)(cn – k), y = anl + b(l + (m + 1)(n + 1)) – (l + m + 1)(cn – k), z = al(n + 1) + bm(n + 1) – (l + m)(cn – k).
(3.9) (3.10) (3.11)
The equation of FP is bx + (a – b)z = ay.
(3.12)
and the condition that F' lies on FP is once again Equation (2.5). It is now a matter of some heavy technical work to establish that ef = EF^E'F' lies on the axis ab bc. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 3
Article: CJB/243/2012
Some properties of a Special Hexagon Christopher Bradley Abstract: The hexagon ABCDEF is such that AD, BE, CF are concurrent at a point U. Given the quadrilaterals ABDE, ACDF, BCEF the diagonal points (other than U) are labelled V, W, V', W', V'', W''. It is proved that these six points lie on four lines with three points on each line and such that each of the six points lies on two of the lines.
V'
X W' To V'
W W'' Y
To Z V'' A
B
F U C D V
E
Figure The diagonal points of the main quadrilaterals of the hexagon 1. Co-ordinates of points We take ABC as triangle of reference and use homogeneous projective co-ordinates. The vertex of perspective of triangles ABC and DEF is labelled U and has co-ordinates U(1, 1, 1). It follows there are parameters p, q, r so that the points D, E, F have co-ordinates D(1 + p, 1, 1), E(1, 1 + q, 1), F(1, 1, 1 + r). 2. Quadrilateral ABDE The equation of AB is z = 0 and the equation of DE is 1
qx + py = (pq +p + q)z.
(2.1)
The co-ordinates of V = AB^DE are therefore V(– p, q, 0). The equation of the line BD is x = (p + 1)z and the equation of the line AE is y = (q + 1)z. It follows that the co-ordinates of W = BD^AE are W(p + 1, q + 1, 1). 3. Quadrilateral ACDF The equation of AC is y = 0 and the equation of DF is rx + pz = (pr + p + r)y = 0.
(3.1)
The co-ordinates of V'= AC^DF are therefore V'(– p, 0, r). The equation of the line AF is z = (r + 1)y and the equation of the line CD is x = (p + 1)y. It follows that the co-ordinates of W' = AF^CD are W'(p + 1, 1, r + 1). 4. Quadrilateral BCEF The equation of BC is x = 0 and the equation of EF is ry + qz = (qr + q + r)x.
(4.1)
The co-ordinates of W'' = BC^EF are therefore W''(0, – q, r). The equation of the line BF is z = (r + 1)x and the equation of the line CE is y = (q + 1)x. It follows that the co-ordinates of V'' are V''(1, q + 1, r + 1). 5. The four lines It may now be checked that the equation of the line WW'W'' is (p + 1)(ry + qz) = (qr + q + r)x.
(5.1)
It may also be checked that the equation of the line VW'V'' is (r + 1)(qx + py) = (pq + p + q)z.
(5.2)
Further it may be checked that the equation of the line VV'W'' is qrx + rpy + pqz = 0.
(5.3)
And finally it may be checked that the equation of the line WV'V'' is (q + 1)(rx + pz) = (pr + p + r)y. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
(5.4)
3
Article: CJB/244/2012
When two Triangles in Perspective create Conics Christopher Bradley Abstract: When triangles ABC and DEF are in perspective through a point I a conic does not normally pass through A, B, C, D, E, and F. It is proved here if the position of F is moved along CI a position may always be found for F so that a conic passes through all six points. When this is so other conics may be drawn through suitably created points.
J
U K
F A V B
M L R N PI Q
S W
C
D G
T
Figure 1
E H
1. Triangles in perspective inscribed in a conic We take two triangles ABC and DEF in perspective through a point I. We assume ABC is the triangle of reference and that I has co-ordinates (1, 1, 1). Homogeneous projective co-ordinates are used throughout. Given the vertex of perspective is I there are thus three constants p, q, r so that the co-ordinates of D, E, F are D(1 + p, 1, 1), E(1, 1 + q, 1), F(1, 1, 1 + r). Any conic through A, B, C has an equation of the form fyz + gzx + hxy = 0,
(1.1)
where f, g, h are constants to be determined. For the conic to pass through D and E we find f, g, h may be chosen as f = – q(p + 1), g = – p(q + 1), h = pq + p + q. (1.2) The condition now that F lies on this conic turns out to be r = – pq/(2pq + p + q).
(1.3)
What this means geometrically is that F has to be moved into a position that depends on p and q so that it has co-ordinates F(2pq + p + q, 2pq + p + q, pq + p + q). (1.4) The equation of the conic is then q(p + 1)yz + p(q + 1)zx = (pq + p + q)xy.
(1.5)
2. Conic LMNPQR Once the condition (1.3) of Section 1 has been satisfied it is found that two other conics exist, passing through six points of intersection of various pairs of lines. In this section we determine a conic interior to conic ABCDEF. We have the equation of AC is y = 0, that of BD is x = (p + 1)z, that of CE is y = (q + 1)x, DF is qx + p(q + 1)y = (2pq + p + q)z that of EA is y = (q + 1)z and that of FB is (pq + p + q)x = (2pq + p + q)z. The co-ordinates of the six points are: L = AE^FB: x = 2pq + p + q, y = (q + 1)(pq + p + q), z = pq + p + q. M = FB^AC: x = 2pq + p + q, y = 0, z = pq + p + q. N = AC^BD: x = p + 1, y = 0, z = 1. P = BD^CE: x = p + 1, y = (p + 1)(q + 1), z = 1. Q = CE^DF: x = 2pq + p + q, y = (q + 1)(2pq + p + q), z = p(q + 1)2 + q. R = DF^EA: x = 1 – pq, y = q + 1, z = 1. 2
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
The general conic has an equation of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0.
(2.7)
The conic passing through the six points L, M, N, P, Q, R has u = 2(pq + p + q), v = – 2p2, w = 2(p + 1)(2pq + p + q), f = – pq(p + 1), g = – p2(q + 1) – 2p(2q + 1) – 2q, h = p(pq + p + q).
(2.8) (2.9) (2.10) (2.11) (2.12) (2.13)
3. Conic UVWGHK In this section we determine a conic UVWGHK that is exterior to the conic ABCDEF. We have the equation of AB is z = 0, that of BC is x = 0, that of CD is x = (p + 1)y, that of DE is qx + py = (pq + p + q)z, that of EF is q(p + 1)x + py = (2pq + p + q)z = 0 and that of FA is (pq + p + q)y = (2pq + p + q)z. The co-ordinates of the six points are: U = AB^EF: x = – p, y = q(p + 1), z = 0. V = FA^BC: x = 0, y = 2pq + p + q, z = pq + p + q. W = AB^CD: x = p + 1, y = 1, z = 0. G = BD^DE: x = 0, y = pq + p + q, z = p. H = CD^EF: x = (p + 1)(2pq + p + q), y = 2pq + p + q, z = p2q + 2pq + p + q. K = DE^FA: x = p2q + 2pq + p + q, y = 2pq + p + q, z = pq + p + q.q
(3.1) (3.2) (3.3) (3.4) (3.5) (3.6)
The conic passing through the six points U, V, W, G, H, K is of the form (2.7) with u = – 2q(p + 1)(pq + p + q), v = 2p(p + 1)(pq + p + q), w = 2(p + 1)(pq + p + q)(2pq + p + q), f = – (p + 1)(p2(q2 + 4q + 2) + pq(2q + 3) + q2), g = – p(p2q(q + 1) + p(q – 1)(2q + 1) + q(q – 1)), h = (pq + p + q)(p2q + p(2q – 1) + q).
(3.7) (3.8) (3.9) (3.10) (3.11) (3.12)
4. The polar line JST The line JST is the polar of I with respect to all three conics. We have J = BC^EF: x = 0, y = (2pq + p + q), z = p. S = FA^CD : x = (p + 1)(2pq + p + q), y = 2pq + p + q, z = pq + p + q. T = AB^DE: x = – p, y = q, z = 0. 3
(4.1) (4.2) (4.3)
And the equation of JST is qx + py = (2pq + p + q)z
(4.4)
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Article: CJB/245/2012
Some Simple Algebraic Results Christopher Bradley Part 1 The conic passing through (1, 0, 0), (0, 1, 0), (0, 0, 1), (u, v, w), (p, q, r). This is of the form fyz + gzx + hxy = 0.
(1.1)
f = pu(qw – rv), g = qv(ru – pw), h = rw(pv – qu).
(1.2)
where
Part 2 Some results about the conic y2 = zx. 1. A point on this conic has the parametric form (1, a, a2), where a may equal 0, 1, ∞. 2. The chord joining A(1, a, a2) and B(1, b, b2) has equation abx – (a + b)y + z = 0. (2.1) 3. The chords joining A and B, and C and D meet at Q with co-ordinates x = a + b – c – d, y = ab – cd, z = abc + abd – acd – bcd. (2.2) 2 4. The point where the chord through P(1, p, p ) and Q meets the conic again has co-ordinates x = (ab – ap – bp – cd + cp + dp)2, y = √(zx), z = (abc + abd – abp – acd – bcd + cdp)2. (2.3) Part 3 The conic yz + zx + xy = 0 1. A point on this conic has parametric form x = s(s – t + 1), y = (1 – t)(s – t + 1) z = s(t – 1). (3.1) Part 4 A conic containing 3 pairs of points in perspective The points are (1, 0, 0), (0, 1, 0), (0, 0, 1), (1 + p, 1, 1), (1, 1 + q, 1), (1, 1, r + 1) – in perspective through (1, 1, 1). The conic is q(p + 1)yz + p(q + 1)zx = (pq + p + q)xy. (4.1) and (r + 1) = (pq + p + q)/(2pq + p + q).
(4.2)
Article: CJB/246/2012
Pascal points Christopher Bradley Abstract: An analysis of a Pascal point is given, using the conic y2 = zx and general points.
V
AC with FB and CD with BE and DF with EA P T
R A
F
K
Z
J
To Y
E
X D
B C
U
S
The Pascal line XYZ and the Pascal point P
1
Q
1. Introduction The conic y2 = zx is given with general points A, B, C, D, E, F having co-ordinates A(1, a, a2) etc. We use the sides AC, FB meeting at X, the sides CD, BE meeting at Y and sides DF, EA meeting at Z to define the Pascal line XYZ. Tangents at A and C meet at S and those at F and B meet at T. Tangents at C and D meet at U and those at B and E meet at V. Tangents at D and F meet at Q and those at E and A meet at R. Then lines ST, UV, QR are concurrent at P the Pascal pole of XYZ. The aim of this article is to provide a general analysis. 2. Points A, B, C, D, E, F and the tangents at those points The conic we have chosen for the general analysis is y2 = zx and the six points on it A, B, C, D, E, F, have co-ordinates A(1, a, a2) etc. The tangent at A has equation 2ay = a2x + x,
(2.1)
with similar equations for the other tangents. Note that the equation of AB is abx + z = (a + b)y.
(2.2)
3. The points X, Y, Z and the Pascal line XYZ X = AC^FB has co-ordinates (x, y, z), where x = a – b + c – f, y = ac – bf, z = abc + acf – abf – bcf. Y = CD^BE has co-ordinates (x, y, z), where x = c – b + d – e, y = cd – be, z = bcd + cde – bce – bde. Z = DF^EA has co-ordinates (x, y, z), where x = a – d + e – f, y = ae – df, z = ade + aef – adf – def.
(3.1)
(3.2) (3.3)
The Pascal line XYZ has equation (abce – abef – acde + acdf + bcdf + bdef)x + (abf – abc – acf + ade – adf + aef + bcd – bce + bcf – bde + cde – def)y + (ac – ae + be– bf – cd + df)z = 0. (3.4) 2
4. Lines ST, UV, QR and the Pascal point P The tangents at A and C meet at S with co-ordinates (1, ½(a + c), ac). The tangents at F and B meet at T with co-ordinates (1, ½(f + b), fb). The equation of ST is (bcf – abc + abf – acf)x + 2(ac – bf)y + (b + f – a – c)z = 0.
(4.1)
The equation of UV is similarly (bce – bcd + bde – cde)x + 2(cd – be)y + (b + e – c – d)z = 0.
(4.2)
The equation of QR is likewise (ade – adf + aef – def)x + 2(df – ae)y + (a + e – d – f)z = 0.
(4.3)
ST, UV, QR concur at the Pascal point P with co-ordinates (x, y, z), where x = ac + be + df – ae – bf – cd, (4.4) y = (1/2)(abc – abf + acf + adf – ade – aef +bde – bcd + bce – bcf – cde + def), (4.5) z = (abce – abef + acdf – acde + bdef – bcdf). (4.6)
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
3
T T X L
K
O
M
Article: CJB/247/2012 U
The C G Conic L
X Christopher Bradley K H Abstract: Through the centroid G lines parallel to the sides are drawn intersecting theMsides in six points. A conic passes through these six points, which we call the G Conic. Its centre is G and we obtain its equation. B S C V A B
S
V
T
L
B
U
G
K
M
S
V
C
The G Conic 1. Analysis The line through G parallel to the line BC has equation 2x = y + z.
(1.1)
This meets CA at M(1, 0, 2) and AB at L(1, 2, 0). Similarly S has co-ordinates S(0, 1, 2), V has co-ordinates V(0, 2, 1), U has co-ordinates U(2, 0, 1), and T has co-ordinates T(2, 1, 0). The equation of the conic through these six points is 2x2 + 2y2 + 2z2 – 5yz – 5zx – 5xy = 0. This conic we term the G Conic and its centre is as expected at G itself.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 1
(1.2)
Article: CJB/248/2012
The H Conic Christopher Bradley Abstract: Lines are drawn parallel to the sides through the orthocentre H of a triangle ABC meeting the sides in six points. The conic through these six points we term the H conic. We obtain its equation and the co-ordinates of its centre X. It is shown that X, H and K, the symmedian point, are collinear. A
U A T T
X
K
O
S
M
C
U
X L
K
B
M
S
V A
H
C
The H Conic
1. The co-ordinates of the six points T U Initially we take the co-ordinates of H to be (u, v, w). The equation of the line through H parallel to the x-axis is (v + w)x = u(y + z). (1.1) L G K M This meets the side CA, y = 0, at the point M(u, 0, v + w) and it meets the side AB, z = 0 at the point L(u, v + w, 0).
B
S C V Similarly the co-ordinates of T are (w + u, v, 0) and those of S are (0, v, w + u). Those of V are (0, u + v, w) and those of U are (u + v, 0, w). 2. The equation of the H Conic
1
The H conic has an equation of the form lx2 + my2 + nz2 + 2fyz + 2gzx + 2hxy = 0.
(2.1)
By substituting in the co-ordinates of the six points L, M, S, T, U, V we find l = 2vw(v + w), m = 2wu(w + u), n = 2uv(u + v), (2.2) 2 2 2 f = – u(u + uv + uw + 2vw), g = – v(v + vw + vu + 2wu), h = – w(w + wu + wv + 2uv). (2.3) For the point H we have u = 1/(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2).
(2.4)
3. The centre X of the H conic The co-ordinates of the centre X of the conic (2.1) are x = mn – gm – hn – f2 + fg + hf, y = nl – hn – fl – g2 + gh + gf, z = lm – fl – gm – h2 + hf + hg.
(3.1) (3.2) (3.3)
Substituting in these equations values of a, b, c and simplifying we find x = (a4 + a2(2b2 + 2c2) – 3b4 – 2b2c2 – 3c4)/(b2 + c2 – a2)2,
(3.4)
with y, z following by cyclic change of a, b, c. It may now be checked that X, H and K(a2, b2, c2) are collinear.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/249/2012
Twin Conics Christopher Bradley Abstract: In a triangle ABC if lines through the symmedian point K are drawn parallel to the sides to meet opposite sides in 6 points, then it is well known that the conic through these six points is a circle called The Triplicate Ratio Circle. If K is exchanged for another point Q and the same construction is carried out then a conic passes through the 6 points which we call the Q conic. If X is the centre of this conic then we show there is a second point Q' also having X as centre and arising from the same construction. If Q is at the centroid G then Q' is also at G. We show that the Triplicate Ratio Circle is twinned to the O conic and we determine the twin of the H conic. The general point Q is first treated and the equation of the Q conic is obtained as also its centre X. The twin conic, the Q' conic, is also obtained.
A U
T'
A T
U'
L
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O
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U'
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M M' T'
B
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U M
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The Triplicate Ratio Circle and the O Conic 1. Analysis for the Q Conic
B
V'
V
S
S'
We take the point Q to have co-ordinates Q(u, v, w). The equation of the line through Q parallel A to the x-axis is (v + w)x = u(y + z). (1.1) T U This meets the side CA, y = 0, at the point M(u, 0, v + w) and it meets the side AB, z = 0, at the point L(u, v + w, 0) L G Similarly the co-ordinates of T on AB are T(w + u, v, 0) and those of S on BC are S(0, v, w + u).M 1 B
V
S
C
Those of V on BC are (0, u + v, w) and those of U on CA are U(u + v, 0, w). We take the equation of a conic to be of the form lx2 + my2 + nz2 + 2fyz + 2gzx + 2hxy = 0
(1.2)
By substituting the co-ordinates of the 6 points L, M, S, T, U, V into (1.2) we find that a conic does indeed pass through the six points with l = 2vw(v + w), m = 2wu(w + u), n = 2uv(u + v), (1.3) 2 2 2 f = – u(u + uv + wu + 2vw), g = – v(v + vw + uv + 2wu), h = – w(w + wu + vw + 2uv). (1.4) The co-ordinates of the centre X of the conic with equation (1.2) are (x, y, z), where x = mn – gm – hn – f2 + fg + hf, y = nl – hn – fl – g2 + gh + fg, z = lm – fl – gm – h2 + hf + gh.
(1.5) (1.6) (1.7)
After normalization the point X has co-ordinates (x, y, z), where x = (1/k)(u(2vw + uv + wu – u2)), y = (1/k)(v(2wu + vw + uv – v2)), z = (1/k)(w(2uv + wu + vw – w2)),
(1.8) (1.9) (1.10)
where k = 2(uv + vw + wu) – (u2 + v2 + w2).
(1.11)
Here we have assumed u + v + w = 1. 2. The point Q' and the Q' conic, also centre X We assert that X is the centre of QQ' and if that is the case the Q' has co-ordinates (p, q, r) where p = (1/k)u(v + w – u), (2.1) q = (1/k)v(w + u – v), (2.2) r = (1/k)w(u + v – w). (2.3) It is a slightly laborious task to show that the conic with p, q, r instead of u, v, w does indeed have centre X. 3. The Triplicate Ratio Circle and its twin the O conic If Q is the point K with co-ordinates u = a2/(a2 + b2 + c2), v = b2/(a2 + b2 + c2), w = c2/(a2 + b2 + c2).
2
(3.1)
T'
U A
T
B
U' U' T L from (2.1) then see that K' has co-ordinates proportional to X – (2.3) we M 2 2 2 2 2 2 2 2 2 2 2 2 O (3.2) K (a (b + c – a ), b (c + a – b ), c (a + b – c )) L' M' T' These will be recognized as the co-ordinates of the circumcentre O. X, of course, U is the centre of the Brocard circle. See the initial diagram above. M' H' V' S' V C S L' L X K H M 4. The G conic is its own twin
If u = v = w = 1/3 the normalized values of p, q, r are also 1/3 showing G' = X = G and that the G conic is its own twin. C S B V' V S' A T
L
U
G M
B
S
V
C
The G Conic 5. The H Conic and its twin Their diagram is below. For the point H the unnormalised values of u, v, w are u = 1/(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2).
(5.1)
Using equation (1.8) we find the x-co-ordinate of the H Conic is x = (a4 + a2(2b2 + 2c2) – 3b4 – 2b2c2 – 3c4)/(b2 + c2 – a2)2,
(5.2)
with y, z following by cyclic change of a, b, c. The point H', as seen above is such that HX = XH'. Points H and O are the only points Q such that QXQ' passes through the symmedian point K. 3
A U
T'
A T
U'
L
X
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L'
K
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T
M M' T'
V
V'
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B
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L'
V'
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'
U M' K H
S'
M
S
C
The H Conic and its twin A The x-co-ordinate of the point H' has unnormalised co-ordinates ((b2 – c2)2 + 2a2(b2 + c2) – 3a4)/(b2 + c2 – a2). T U
Flat 4, L Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. B
G M
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S
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4
(5.3)
Article: CJB/250/2012
Cevian Circles Christopher Bradley Abstract: The feet of a set of Cevians through a point S are three points through which a circle may be drawn. The other three points where the circle cuts the sides are also the feet of a set of Cevians through a point T. The co-ordinates of T are obtained in terms of those of S. When S is the centroid G then T is the orthocentre. There do not appear to be any other important circles deserving the title Cevian Circles.
A
W V T N S
B
U
L
M
C
A Cevian Circle 1. The Cevian Circle The equation of a circle is of the form a2yz + b2zx + c2xy – (ux + vy + wz)(x + y + z) = 0,
(1.1)
where u, v, w are constants to be determined. Take S to be the point with co-ordinates S(p, q, r). Then the feet of the Cevians are L(0, q, r), M(p, 0, r), N(p, q, 0). Substituting these co-ordinates in Equation (1.1) we find u = (p(q + r)(b2r(p + q) + c2q(p + r)) – a2qr(p + q)(p + r))/(2p(p + q)(p + r)(q + r)), (1.2) 2 2 2 v = a (qr(p + q)(p + r) + p(q + r)(b r(p + q) – c q(p + r))/(2q(p + q)(p + r)(q + r)), (1.3) 1
w = a2qr(p + q)(p + r) – p(q + r)(b2r(p + q) – c2q(p + r))/(2r(p + q)(p + r)(q + r)).
(1.4)
The equation of the Cevian circle is now obtained by inserting these values of u, v, w into Equation (1.1). 2. The co-ordinates of the second Cevian point T The other points of intersection U, V, W of the Cevian circle may now be obtained and these of course turn out to be the feet of the Cevian point T with co-ordinates (x, y, z), where x = 1/(p(q + r)(b2r(p + q) + c2q(p + r)) – a2qr(p + q)(p + r)), (2.1) 2 2 2 y = 1/(a qr(p + q)(p + r) – p(q + r)(b r(p + q) – c q(p + r))). (2.2) 2 2 2 z = 1/(a qr(p + q)(p + r) + p(q + r)(b r(p + q) – c q(p + r))). (2.3) It may be checked that when (p, q, r) is the centroid G(1, 1, 1), then T is the orthocentre H with co-ordinates H(1/(b2 + c2 – a2), 1/(c2 + a2 – b2), 1/(a2 + b2 – c2)) and the Cevian circle is the ninepoint circle. Unfortunately there does not appear to be any other known Cevian circle in which both S and T have well known co-ordinates except possibly when S is the symmedian point K(a2, b2 c2) and then T has x-co-ordinate x = 1/(b4 + c4 + a2b2 + b2c2 + c2a2 – a4), (2.4) with y- and z- co-ordinates obtained by cyclic change of a, b, c. Kimberling’s list of centres.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
This point is not in
Article: CJB/251/2012
The Four Brocard Triangles Christopher Bradley Abstract: The six circles used to construct the Brocard points are analysed to produce the First and Second Brocard triangles situated on the circumference of the Brocard circle and similar constructions produce the Third and Fourth Brocard triangles which turn out to be the Isogonal Conjugates of the first two.
A
R O
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R' C
THE BROCARD TRIANGLES
1. The Brocard circle and its isogonal conjugate locus The equation of the Brocard circle is b2c2x2 + c2a2y2 + a2b2z2 – a4yz – b4zx – c4xy = 0
(1.1)
It is known, of course that this has OK as a diameter, where O is the circumcentre of triangle ABC and K is the symmedian point. Also, the two Brocard points W(1/b2, 1/c2, 1/a2) and W'(1/c2, 1/a2, 1/b2) lie on the circle, with the chord WW' perpendicular to OK. The isogonal conjugate locus of the Brocard circle ha equation a2y2z2 + b2z2x2 + c2x2y2 – xyz(a2x + b2y + c2z).
(1.2)
This contains the orthocentre H, the centroid G and again, the pair of Brocard points W and W'. 2. The six defining circles The six defining circles of the Brocard points are: BCW: – b2x2 + (c2 – b2)xy + a2yz = 0, CAW: b2zx – c2y2 + (a2 – c2)yz = 0, ABW: – a2z2 + c2xy + (b2 – a2)zx = 0, BCW': – c2x2 + (b2 – c2)zx + a2yz = 0, CAW': b2zx – a2y2 + (c2 – a2)xy = 0, ABW': – b2z2 + c2xy + (a2 – b2)yz = 0.
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
3. The Second Brocard triangle PQR The point P = CAW^ABW' has co-ordinates P(b2 + c2 – a2, b2, c2) and it is easily checked that P lies on the Brocard circle. The other two vertices of the Second Brocard triangle have coordinates Q(a2, c2 + a2 – b2, c2) and R(a2, b2, a2 + b2 – c2). 4. The Fourth Brocard triangle P'Q'R' The point P' = ABW^CAW' has co-ordinates P'(a2, b2 + c2 – a2, b2 + c2 – a2) and it is easily checked that P' lies on the isogonal locus of the Brocard circle. In fact, P' is the isogonal conjugate of P. The other two vertices of the Fourth Brocard triangle have co-ordinates Q'(c2 + a2 – b2, b2, c2 + a2 – b2) and R'(a2 + b2 – c2, a2 + b2 – c2, c2) and these are the isogonal conjugates of Q and R respectively. 5. The First Brocard triangle LMN
1
The equation of BW is x/a2 = z/b2,
(5.1)
and the equation of CW' is x/a2 = y/c2.
(5.2)
The point L = BW^CW' has co-ordinates L(a2, c2, b2). The other two vertices M and N of the First Brocard triangle have co-ordinates M(c2, b2, a2) and N(b2, a2, c2). 6. The Third Brocard Triangle The equation of CW is x/c2 = y/b2,
(6.1)
and the equation of BW' is x/b2 = z/c2.
(6.2)
The point L' = CW^BW' has co-ordinates L'(b2c2, b4, c4). This point lies on the isogonal locus of the Brocard circle and indeed L' is the isogonal conjugate of L. The other two vertices M' and N' of the Third Brocard triangle have co-ordinates M'(a4, a2c2, c4) and N'(a4, b4, a2b2).
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP.
2
Article: CJB/252/2012
Odds and Ends Christopher Bradley
Part 1: Equation of Circle from radius and centre If the radius is r and the centre is (l, m, n) with l + m + n = 1 then the equation is ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(1.1)
where u = a2mn + b2n(l – 1) + c2m(l – 1) + r2 with v, w following from (1.2) by cyclic change of a, b, c and l, m, n, and 2f =a2(2mn – m – n + 1) + b2l(2n – 1) + c2l(2m – 1) + 2r2,
(1.2)
(1.3)
with 2g, 2h following from (1.3) by cyclic change of a, b, c and l, m, n. This may be checked by showing (v + w – 2f)/a2 = (w + u – 2g)/b2 = (u + v – 2h)/c2 ( = – 1).
(1.4)
Part 2: Change of Brocard points and the construction of the 2nd and 4th B-Triangles In the Brocard construction instead of using the Brocard points we use another isogonal pair K(a2, b2, c2) and G(1, 1, 1) – labelled V, V' in the diagram below. Circle BCV has equation 3b2c2x2 – c2(c2 + a2 – 2b2)xy – b2(a2 + b2 – 2c2)zx – a2(a2 + b2 + c2)yz = 0.
(2.1)
Circles CAV and ABV have equations which may be written down from (2.1) by cyclic change of x, y, z and a, b, c. Circle BCV' has equation (a2 + b2 + c2)x2 + (a2 + b2 – 2c2)xy + (c2 + a2 – 2b2)zx – 3a2yz = 0.
(2.2)
Circles CAV' and ABV' have equations which may be written down from (2.2) by cyclic change of x, y, z and a, b, c. 1
A
P R
Q'
V
V' P'
Q R'
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The B-Conic The Second B-Point P = CAV^ABV' has co-ordinates (x, y, z), where x = 3a2|(2a6 + 3a4(b2 – 2c2) – 9c2a2(b2 + c2) – (b2 + c2)3), y = (a2 + b2 + c2)2(2a4 + a2(c2 – 8b2) – (b2 + c2)2), z = 9c2a2(2a4 + a2(c2 – 8b2) – (b2 + c2)2). 2
(2.3) (2.4) (2.5)
The other Second B-Points Q = ABV^BCV' and R = BCV^CAV' have co-ordinates that may be written down from (2.3) – (2.5) by cyclic change of x, y, z and a, b, c.] The B-Conic V, V', P, Q, R may now be drawn and also the locus of the isogonal conjugates of the B-Conic. Unfortunately the equations of the conic and the locus are too lengthy and technically complicated to record with entire accuracy. The Fourth B-Point P' = ABV^CAV' can now be shown to have co-ordinates that are the isogonal conjugates of P and similarly for Q' and R'. It may be shown that the construction of the First and Third B-triangles also works – however, the First and Second B-triangle vertices only lie on the same conic when V, V' coincide with the Brocard points and then the conic is the Brocard circle.
3
Article: CJB/253/2012
Triangles in Perspective – the Whole Truth Christopher Bradley Abstract: Triangles ABC and 123 are in perspective if A1, B2, C3 are concurrent at a point X commonly called the perspector. It then follows that the points AB^12, BC^23, CA^31 are collinear. The line is called the Desargues’ axis of perspective and nowadays is often called the perspectrix. That is all that is taught to elementary classes. However, it is in fact far from the end of the matter. If we define the cross-over triangle to have vertices B1^A2, C2^B3, A3^C1 then we show that B1^A2, C2^B3 and CA^31 are collinear as are B1^A2, A3^C1and BC^23 and A3^C1, C2^B3 and AB^12. In other words there is not just an axis of perspective but a complete quadrilateral of perspective. But even that is not all. We prove also that the remaining six points A2^B3, B1^C2, B3^C1, C2^A3, A3^B1, C1^A2 lie on a conic. This last result I have not yet located in past literature. 1. Vertices and sides of the triangles and the axis of perspective We use homogeneous projective co-ordinates throughout, with ABC as triangle of reference and X, the perspector, having co-ordinates X(1/3, 1/3, 1/3). The points 1, 2, 3 lie on AX, BX, CX respectively and so have co-ordinates of the form 1(1 – 2k, k, k), 2(h, 1 – 2h, h), 3(g. g. 1 – 2g), (1.1) where for the sake of generality none of k. h. g is 1/2 or 1/3 and no pair of them have equal values. AB has equation z = 0 and 12 has equation k(3h – 1)x + h(3k – 1)y + (3hk – 2k – 2h + 1)z = 0.
(1.2)
This pair of corresponding sides meet at AB^12 with co-ordinates (h(1 – 3k), k(3h – 1), 0). Similarly BC^23 has co-ordinates (0, g(1 – 3h), h(3g – 1)) and CA^12 has co-ordinates (g(3k – 1), 0, k(1 – 3g)). These three points lie on Desargues’ axis of perspective with equation kx/(3k – 1) + hy/(3h – 1) + gz/(3g – 1) = 0 (1.3)
1
AC 13
1
C1 A2
A3 B1 A
A3 C1
B1 A2 BC 23
2 A2 B3
B1 C2 B
3 B3 C1
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C C2 A3
X
AB 12
Perspective triangles, quadrilateral and conic
2
2. The six joins of vertices of ABC and 123 other than A1, B2, C3 These have equations as follows: A2: hy = (1 – 2h)z. B3: gz = (1 – 2g)x. C1: kx = (1 – 2k)y. A3: gz = (1 – 2g)y. B1: kx = (1 – 2k)z. C2: hy = (1 – 2h)x.
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
3. The vertices of the cross-over triangle These are: B1^A2: C2^B3: A3^C1:
(h(1 – 2k), k(1 – 2h), hk). (gh, g(1 – 2h), h(1 – 2g)). (g(1 – 2k), gk, k(1 – 2g)).
(3.1) (3.2) (3.3)
B1^A2 is the point of intersection of the lines joining Vertex 1 of one triangle to vertex 2 of the other and vice-versa. Hence the name given is ‘cross-over’. It may now be proved by showing the 3 x 3 determinant of coefficients of the three co-ordinates involved is zero that the following sets of points are collinear: (1) B1^A2, A3^C1, BC^23; (2) C2^B3, B1^A2, CA^31; (3) A3^C1, C2^B3, AB^12; These three lines together with Desargues’ axis hence form a complete quadrilateral of perspective. David Monk (private communication) has pointed out that the reason there are four axes of perspective is that not only ABC and 123 are in perspective, but also 1BC and A23, A2C and 1B3, AB3 and 12C 4. The Conic of Perspective The six points we consider have co-ordinates as follows: A2^B3: (gh, (1 – 2g)(1 – 2h), h(1 – 2g)). B3^C1: (g(1 – 2k), gk, (1 – 2k)(1 – 2g)). C1^A2: ((1 – 2h)(1 – 2k), k(1 – 2h), hk). A3^C2: (gh, g(1 – 2h), (1 – 2g)(1 – 2h)). B1^A3: ((1 – 2g)(1 – 2k), gk, k(1 – 2g)). C2^B1: (h(1 – 2k), (1 – 2k)(1 – 2h), hk). 3
(4.1) (4.2) (4.3) (4.4) (4.5) (4.6)
The intersections of the lines in Section 2 create nine points, three of which are the vertices of the cross-over triangle. The above six points are the remaining six points of the nine. Remarkably they lie on a conic with equation ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0, (4.7) where u = 2k(3k – 1)(2h – 1)(2g – 1)(3gh – 2g – 2h + 1), (4.8) v = 2h(3h – 1)(2g – 1)(2k – 1)(3gk – 2k – 2g + 1), (4.9) w = 2g(3g – 1)(2k – 1)(2h – 1)(3hk – 2h – 2k + 1), (4.10) l = (2k – 1)(3h – 1)(3g – 1)(7ghk – 4gh – 4kg – 4hk + 2g + 2h + 2k – 1), (4.11) m = (2h – 1)(3g – 1)(3k – 1)(7ghk – 4gh – 4kg – 4hk + 2g + 2h + 2k – 1),(4.12) n = (2g – 1)(3k – 1)(3h – 1)(7ghk – 4gh – 4kg – 4hk + 2g + 2h + 2k – 1). (4.13) It does, of course, seem likely that this conic has been discovered before, but I have been unable to find any reference to it in the literature, Because there are four pairs of triangles in perspective (see Section 3) there are in fact 4 Conics. The complete figure is shown in the final figure.
4
A
3 2 B
C 1
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 5
Article: CJB/254/2012
Circles through the midpoints of sides of a Cyclic Quadrilateral Christopher Bradley Abstract: Let ABCD be a cyclic quadrilateral with diameter AC, with centre O and with external diagonal points Q and R and let S, T be the midpoints of AQ, AR and U, V the midpoints of BC, CD. Then circles QUVR and BOSTD exist and their line of centres passes through the midpoint of AO.
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The midpoint circles 1. Introduction 1
Let ABCD be a cyclic quadrilateral with diameter AC, centre O and with external diagonal points Q and R and let S, T be the midpoints of AQ, AR and U, V the midpoints of BC, CD. Then circles QUVR and BOSTD exist and their line of centres passes through the midpoint of AO Proofs are given using Cartesian Co-ordinates centre O with circle ABCD having equation x2 + y2 = 1. (1.1) We take co-ordinates of B, C, D to have standard form x = 2k/(1 + k2), y = (1 – k2)/(1 + k2),
(1.2)
with k = b, c, d respectively. Since A is diametrically opposite C we know its coordinates must be x = – 2c/(1 + c^2), y = – (1 – c2)/(1 + c2). (1.3) The equations of the sides of ABCD are therefore of the form, for BC (b + c)x + (1 – bc)y = (1 + bc),
(1.4)
and similarly for CD with d replacing b. However AB has equation (bc – 1)x + (b + c)y = c – b,
(1.5)
And similarly for AD with d replacing b. The diagonal AC which passes through O has equation (1 – c2)x = 2cy. (1.6)
2. Points Q, R, U, V The diagonal point Q = AB^CD and has co-ordinates
.
x = 2(b + c2d)/((1 + c2)(1 + bd)), y = (1 + 2cd + c2 – bc2d – 2bc – bd)/((1 + c2)(1 + bd)).
(2.1) (2.2)
The diagonal point R = AD^BC has co-ordinates the same as Q but with b and d interchanged. The midpoint of BC is the point U with co-ordinates x = (b2c + bc2 + b + c)/((1 + b2)(1 + c2)), y = (1 – b2c2)/((1 + b2)(1 + c2)).
(2.3) (2.4)
The midpoint of CD is V with the same co-ordinates as U but with d replacing b. 3. The circles RUV and BOD Circles, of course, have equations of the form x2 + y2 + 2gx + 2fy + k = 0, 2
(3.1)
with centre (– g, – f) and radius squared equal to g2 + f2 – k. We find circle RUV to have k = 2(1 + bc)(1 + cd)/((1 + c2)(1 + bd))
(3.2)
and centre W with co-ordinates x = (c2(b + d) + bcd + c + d)/((1 + c2)(1 + bd)), y = (3 + c2 – bd – 3c2bd)/(2(1 + c2)(1 + bd)).
(3.3) (3.4)
And we find circle BOD has k = 0 and centre P with co-ordinates x = (b + d)/(2(1 + bd)) , y = (1 – bd)/(2(1 + bd)).
(3.5) (3.6)
The midpoint M of AO has co-ordinates x = – c/(1 + c2), y = (c2 – 1)/(2(1 + c2)).
(3.7)
It may now be checked that points W, P, M are collinear. 4. More on Circles BOD and RUV Circle BOD has equation (1 + bd)(x2 + y2) = b + d)x + (1 – bd)y.
(4.1)
This meets AD at D and T, where T has co-ordinates x = (c – d)(bc – 1)/((1 + c2)(1 + bd)) (4.2) y = (b + c)(c – d)/((1 + c2)(1 + bd)). (4.3) It may now be checked that this is the midpoint of AR. Similarly S is the midpoint of AQ. Finally it may be shown that circle RUV passes through Q.
Flat 4, Terrill Court, 12-14, Apsley Road BRISTOL BS8 2SP.
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Article: CJB/255/2012
Five Conics and a Polar line Christopher Bradley Abstract: Two triangles inscribed in a triangle are also in perspective. Five Conics are now created and a polar line which is also two Pascal lines. X
R
To X
W A F
B Y
C E
P D
U
Q
Z
CONICS AND POLAR LINE 1
1. Introduction Two triangles inscribed in a triangle are also in perspective. Five Conics are now created and a polar line which is also two Pascal lines. The two triangles are ABC and DEF in the picture. ABC is the triangle of reference and the vertex of perspective is (1, 1, 1). The points D, E, F for the perspective may be chosen as D(1 + p, 1, 1), E(1, 1 + q, 1), F(1, 1, 1 + r). The equation of the conic ABCDE is q(p + 1)yz + p(q + 1)zx = (pq + p + q)xy.
(1.1)
The condition that F lies on the conic is that r = – pq/(2pq + p + q),
(1.2)
and hence the co-ordinates of F have to be adjusted to F(2pq + p + q, 2pq + p + q), pq + p + q). 2. Second Conic This conic is interior to the defining conic in Section 1. It consists of the following six points: AE^FB: (2pq + p + q, (q + 1)(pq + p + q), pq + p + q), (2.1) FB^AC: (2pq + p + q, 0, pq + p + q), (2.2) AC^BD: (p + 1, 0, 1), (2.3) BD^CE: (p + 1, (p + 1)(q + 1), 1) (2.4) CE^DF: (2pq + p + q, (q + 1)(2pq + p + q), p(q + 1)2 + q), (2.5) DF^EA: (1 – pq, q + 1, 1). (2.6) The general equation of a conic has the form ux2 + vy2 + wz2 + 2fyz + 23gzx + 2hxy = 0.
(2.7)
The conic through the above six points has u = 2(pq + p + q), v = – 2p2, w = 2(p + 1)(2pq + p + q), f = – pq(p + 1), g = – p2(q + 1) – 2p(2q + 1) – 2q, h = p(pq + p + q).
(2.8)
3. Third Conic This conic is exterior to the conic in Section 1 but closer to it than the fourth and fifth conics. It consists of the following six points: AB^EF: (– p, q(p + 1), 0), (3.1) FA^BC: (0, 2pq + p + q, pq + p + q), (3.2) AB^CD: (p + 1, 1, 0), (3.3) 2
BC^DE: (0, pq + p + q, p), CD^EF: ((p + 1)(2pq + p + q), 2pq + p + q, pq2 + 2pq + p + q), DE^FA: (p2q + 2pq + p + q, 2pq + p + q, pq + p + q).
(3.4) (3.5) (3.6)
The conic through these six points has an equation of the form (2.7) wit u = – 2q(p + 1)(pq + p + q), v = 2p(p + 1)(pq + p + q), w = 2(p + 1)(pq + p + q)(2pq + p + q), f = – (p + 1)(p2(q2 + 4q + 2) + pq(2q + 3) + q2), g = – p(p2q(q + 1) + p(q – 1)(2q + 1) + q(q – 1)), h = (pq + p + q)(p2q + p(2q – 1) + q). (3.7) 4. The fourth and fifth conics It has of course been checked analytically that conics pass through the following two sets of six points. However, their equations are technically so long and complicated that copying them down could not be carried out without error. So, alas, they are omitted. Conic four passes through the following six points: AC^EF: (p(2q + 1) + q, 0, q(p + 1)), BD^FA: ((p + 1)(pq + p + q), 2pq + p + q, pq + p + q), CE^AB: (1, q + 1, 0), DF^BC: (0, 2pq + p + q, p(q + 1)), EA^CD: ((p + 1)(q + 1), q + 1, 1), FB^DE: (2pq + p + q, p(q + 1)2 + q, pq + p + q).
(4.1) (4.2) (4.3) (4.4) (4.5) (4.6)
The fifth conic passes through the following six points: AB^DF: (– p(q + 1), q, 0), BC^EA: (0, q + 1, 1), CD^FB: ((p + 1)(2pq + p + q), 2pq + p + q, (p + 1)(pq + p + q)), DE^AC: (pq + p + q, 0, q), EF^BD: (p + 1, 1 – pq, 1), FA^CE: (2pq + p + q, (q + 1)(2pq + p + q), (q + 1)(pq + p + q))
(4.7) (4.8) (4.9) (4.10) (4.11) (4.12)
5. Points on the Polar line It is the polar of the point of perspective with respect to all the conics just listed. The following points lie on this polar line, which has equation qx + py = (2pq + p + q)z. (5.1) AB^DE: (– p, q, 0), (5.2) BC^EF: (0, 2pq + p + q, p), (5.3) CD^FA: ((p + 1)(2pq + p + q), 2pq + p + q, pq + p + q), (5.4) AE^BD: (p + 1, q + 1, 1) (5.5) 3
BF^CE: (2pq + p + q, (q + 1)(2pq + p + q), pq + p + q), DF^CA: (2pq + p + q, 0, q),
FLAT 4, TERRILL COURT, 12-14, APSLEY ROAD, BRISTOL BS8 2SP.
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(5.6) (5.7)
Article: CJB/255/2012
Two Quadrangles in Perspective in a Conic Christopher Bradley Abstract: Two Quadrangles in Perspective in a Conic are drawn. We prove their six diagonal points lie on a conic.
The diagonal points of the quadrangle lie on a conic 1. Introduction Two quadrangles ABCD and EFGH are in perspective and lie in a conic. We prove their diagonal points are co-conic. 1
We suppose the conic has equation y2 = zx with A(a2, a, 1) etc., so the equation of AB is x – (a + b)y + abz = 0. (1.1) Let the vertex of perspective be (0, 1, 0). Then E has co-ordinates ( a2, – a, 1) and the equation of EF is x + (a + b)y + abz = 0. (1.2) 2. The sides of the quadrilateral and the diagonal points AB has equation (1.1) and CD has equation x– (c + d)y + cdz = 0.
(2.1)
The diagonal point AB^CD therefore has co-ordinates x = abc + abd – acd – bcd, y = ab – cd, z = a + b – c – d.
(2.2)
The diagonal points AC^BD may be obtained from (2.2) by interchange of b and c. The diagonal point AD^BC may be obtained from (2.2) by interchange of b and d. The diagonal points of EFGH may be obtained from those of ABCD by change of a, b, c, d to – a, – b, – c. – d. 3. The conic containing the diagonal points This is of the form ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0, where we find: u = a3b + . + . – 2a2b2 – . – . – a2bc – . – . + 12 abcd. v = 4a3bcd + . + . – 4a2b2c2 – . – . f = 0, h = 0, g = v/4, w = a3b3c2 + . + . – 2a3b3cd – . – . – a3b3c2d – . –. + 12 a2b2c2d2 with symmetrical terms left for the reader to complete.
Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
(3.1)
Article: CJB/2012/257
Quadrangles in Perspective Part II The three lines Christopher Bradley Abstract: Two quadrangles are drawn in a conic and are in perspective. Three lines emerge with six points of intersection on each line. They are like Pascal lines but appropriate to Quadrilaterals in perspective and not to pairs of triangles. It is difficult to be sure but they may be a new discovery.
H
B C
A J
E G
D F
The three blue lines 1. Introduction ABCD and EFGH are two quadrangles inscribed in a conic (not just a circle as in the diagram). Various points of intersection to be detailed later are found to exist, six points of intersection on
1
each of three special lines. These lines are similar to Pascal lines but are not related to them and are only three in number. We found them something of a surprise. We take the conic to have equation y2 = zx,
(1.1)
and the vertex of perspective to be J(0, 1, 0). These choices allow us to choose the co-ordinates of A to be A(a2, a, 1) etc and those of E to be (a2, – a, 1), etc. Parameters b, c, d are used for other pairs of points in the perspective. The equation of BC is found to be x – (b + c)y + abz = 0,
(1.2)
and similarly for AD. These intersect at the point BC^AD with co-ordinates BC^AD: (ad(b + c) – bc(a + d), ad – bc, a – b – c + d). 2. Other points of intersection The co-ordinates of other points of intersection may be deduced by change of co-ordinates to a, b, c, d, – a, – b, – c, – d as appropriate. BE^GD: (acd + bcd – abc – bcd, ab – cd, a – b – c + d), CE^FD: (acd + abd – abc – bcd, ac – bd, a – b – c + d), AG^HB: (acd + abd – abc – bcd, bd – ac, a – b – c + d), AF^HC: (acd + abd – abc – bcd, cd – ab, a – b – c + d), BC^AD: (adb + acd – abc –bcd, cad - bc, a – b – c + d), FG^HE: (abc + abd – acd – bcd, cd – ab, a + b – c – d).
(2.1) (2.2) (2.3) (2.4) (2.5) (2.6)
These six points lie on the line with equation (a – b – c + d)x + (bcd + abc – abd – acd)z = 0.
(2.7)
If you write (ADFG/BCEH) you can read off the points of intersection following a fairly obvious pattern. And there are two other lines in which points A, B, C, D, and E, F, G, H are permuted. If the above line is designated by BC^AD then the others are designated by CD^BA and CA^BD. The other lines involve the following sets of 6 points: (AB^CD, AH^FC, AG^FD, EF^GH, AG^FD, GB^ED), (AH^BG, AF^GD, GE^HF, BD^AC, BE^HC, DE^FC). Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 2
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Article: CJB/2012/258
The Miquel Cyclic Quadrilateral Christopher Bradley
A N
D R P
M
L Q S C K
B
Abstract: Let ABCD be a cyclic quadrilateral with points L, M, N on the line segments AB, CD, DA respectively. Circles ANL and DMN meet at N and a point P (which we assume does not lie on AC or BD – otherwise adjust L, M, N). Circle BPL now meets BC at K. We now prove that circle CPM passes through K. We thus have a Miquel point P with four circles through it. The Miquel point is not automatic since the point K has to arise naturally. Circles ANL and CMK meet at P and another point R that we show lies on AC. Circles DMN and BLK meet at P and another point S that we prove lies on BD. We then show circle PRS passes through Q the intersection of AC and BD. 1. Setting up a co-ordinate system and the key points A, B, C, D and the circle ABCD We use Areal co-ordinates throughout. Let A have co-ordinates A(1, 0, 0) and D have coordinates D(0, 0, 1). We suppose that AB and DC meet at (0, 1, 0) so that AB has equation z = 0 1
and DC has equation x = 0. We choose L, M, N to have co-ordinates L(l, 1 – l, 0) since it lies on z = 0, M(0, m, 1 – m) since it lies on x = 0 and N(1 – n, 0, n) since it lies on y = 0. The circle ABCD (since it passes through A and D) has equation a2yz + b2zx + c2xy + vy(x + y + z) = 0 (1.1) for some value of v. It meets z = 0 again at B(– v, c2 + v, 0), and it meets x = 0 again at C(0, a2 + v, – v). 2. The Miquel circles and the points P and K We can now find the equation of circle ALN, which has the form a2yz + b2zx + c2xy + (px + qy + rz)(x + y + z) = 0,
(2.1)
with p = 0, q = – c2l, r = b2(n – 1). Similarly circle DMN has an equation of the form (2.1) with p = – b2n, q = a2(m – 1), r = 0. These two circles meet not only at N but at the Miquel point P with co-ordinates (x, y, z), where x = a2(a2m(m – 1) + b2(m – 1)(n – 1) + c2lm), (2.2) 2 2 2 2 y = b (a mn + b n(n – 1) + c (l – 1)(n – 1)), (2.3) 2 2 2 2 z = c (a (l – 1)(m – 1) + b nl + c l(l – 1)). (2.4) The third circle BLP has an equation of the form (2.1) with p = (c2 + v)(l – 1), q = lv, r = – (a2m(c2 + v) + b2v(n – 1))/c2.
(2.5)
This circle now meets BC, which has the equation (c2 + v)x + vy + (a2 + v)z = 0,
(2.6)
at the point K with co-ordinates (x, y, z), where x = – a2(a2(c2(l + m – 1) + mv) + v(b2(n – 1) + c2(l – 1))), y = a4(c2(l + m – 1) + mv) – a2(b2(c2 – v(n – 1)) + c2(c2(l + m – 1) – v(l – m – 1))) – c2v(b2n + c2(l – 1)), z = c2(a2(c2(l + m – 1) + mv) + v(b2n + c2( l – 1))). Finally the circle CMP has the form (2.1) with p = (a2c2(l – 1) + b2nv + c2v(l – 1))/a2, q = v(1 – m), r = – m(a2 + v).
(2.7) (2.8) (2.9)
(2.10)
and the important result is that circle CMP passes through K. This provides four Miquel circles intersecting at P and only passing through one point on each of the sides AB, BC, CD, DA.
2
3. The points R, S, Q and the circle PQRS The equation of the diagonal AC is vy + (a2 + v)z = 0,
(3.1)
and the equation of the diagonal BD is (c2 + v)x + vy = 0.
(3.2)
These meet at the point Q with co-ordinates Q(v(a2 + v), – (a2 + v)(c2 + v), v(c2 + v)). The diametrically opposite circles ANL and CMP meet at P and again at R with co-ordinates x = a2(a2(c2 l + v) + b2v(n – 1) + c2 lv + v2), (3.3) 2 2 2 2 2 y = – (a + v)(a c (l – 1) + v(b n + c ( l – 1))), (3.4) 2 2 2 2 z = v(a c (l – 1) + v(b n + c ( l – 1))). (3.5) It may be checked that R lies on the diagonal AC. The diametrically opposite circles BMN and BLP meet at P and again at S with co-ordinates x = – v(a2m(c2 + v) + b2v(n – 1)), (3.6) 2 2 2 2 y = (c + v)(a m(c + v) + b v(n – 1)), (3.7) 2 2 2 2 2 z = – c (a (c + v)(m – 1) + v(b n – c – v)). (3.8) It may be checked that S lies on the diagonal BD. The circle QRS has an equation of the form (2.1) with p, q, r given by p = ((c2 + v)(a2c2(l – 1) + v(b2n + c2(l – 1))))/(a2c2 – v2), q = (v(a2(c2(l – m + 1) – v(m – 1)) + c2lv))/(a2c2 – v2), r = ((a2 + v)(a2m(c2 + v) + b2v(n – 1)))/(v2 – a2c2). It may now be checked that the Miquel point P lies on circle QRS.
3
(3.9) (3.10) (3.11)
Article: CJB/2012/259
Two Triangles and an Eight Point Conic Christopher Bradley Abstract: Let ABC and TUV be two triangles with L, M, N the points BC^UV, CA^VT, AB^TU respectively Suppose adjustments are made so that L, M, N are the feet of Cevians of both triangles. Then we prove a conic passes through A, B, C, T, U, V and the two Cevian points.
T A
M N S
P
U B
L
C
To V
V
The two triangles and the conic 1
1. The triangles and their sides We take ABC to be the triangle of reference and the Cevian point P for the triangle ABC to have co-ordinates P(p, q , r). It follows that the points L, M, N have co-ordinates L(0, q, r), M(p, 0, r) and N(p, q, 0). The sides of the triangle TUV pass through L, M, N and have equations: VT: my + pz – rx = 0, (1.1) TU: nz + qx – py = 0, (1.2) UV: lx + ry – qz = 0, (1.3) for some constants l, m, n. It is now possible to determine the co-ordinates of the points T, U, V which are: T: (mn + p2, nr + pq, pr – mq), U: (pq – nr, nl + q2, lp + qr), V: (mq + pr, qr – lp, lm + r2).
(1.4) (1.5) (1.6)
2. The condition that TL, UM, VN meet in a (Cevian) point S The equation of TL is (mq2 + nr2)x – (mn + p2)(ry – qz) = 0.
(2.1)
(ln + q2)(rx – pz) + (lp2 + nr2)y = 0.
(2.2)
(lm + r2)(py – qx) + (lp2 + mq2)z = 0.
(2.3)
The equation of UM is
The equation of VN is
These three lines are concurrent if and only if (lmp2q2 + mnq2r2 + nlr2p2 + p2q2r2)(lp2 + mq2 + nr2 + lmn) = 0.
(2.4)
Either one bracketed term must vanish or the other or possibly even both. When they meet the point of concurrence S has co-ordinates (x, y, z), where x = (lp – qr)(mn + p2)(nr – pq), y = (mq + rp)(nr + pq)(nl + q2), z = lm(n2r2 + p2q2) + 2lmnr2p2 + r2(n2r2 + p2q2) + 2mn q2r2. 3. The 8 pt conic It transpires that the conic with equation 2
(2.5) (2.6) (2.7)
fyz + gzx + hxy = 0,
(3.1)
f = q(mn+ p2)(nr – pq), g = p(nl + q2)(nr + pq), h = n(lp + qr)(mq – rp).
(3.2) (3.3) (3.4)
with
passes through all eight points A, B, C, T, U, V, P, S provided both bracketed terms in equation (2.4) vanish. Essentially this means that given triangle ABC and the Cevian point P, there is one and only one triangle TUV with the properties required.
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